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knowrohit07
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gita_dataset

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7618-7621
What is the probability that (i) all the five cards are spades (ii) only 3 cards are spades (iii) none is a spade 5
1
7619-7622
(ii) only 3 cards are spades (iii) none is a spade 5 The probability that a bulb produced by a factory will fuse after 150 days of use is 0
1
7620-7623
(iii) none is a spade 5 The probability that a bulb produced by a factory will fuse after 150 days of use is 0 05
1
7621-7624
5 The probability that a bulb produced by a factory will fuse after 150 days of use is 0 05 Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use
1
7622-7625
The probability that a bulb produced by a factory will fuse after 150 days of use is 0 05 Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use 6
1
7623-7626
05 Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use 6 A bag consists of 10 balls each marked with one of the digits 0 to 9
1
7624-7627
Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use 6 A bag consists of 10 balls each marked with one of the digits 0 to 9 If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0
1
7625-7628
6 A bag consists of 10 balls each marked with one of the digits 0 to 9 If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0 7
1
7626-7629
A bag consists of 10 balls each marked with one of the digits 0 to 9 If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0 7 In an examination, 20 questions of true-false type are asked
1
7627-7630
If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0 7 In an examination, 20 questions of true-false type are asked Suppose a student tosses a fair coin to determine his answer to each question
1
7628-7631
7 In an examination, 20 questions of true-false type are asked Suppose a student tosses a fair coin to determine his answer to each question If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'
1
7629-7632
In an examination, 20 questions of true-false type are asked Suppose a student tosses a fair coin to determine his answer to each question If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false' Find the probability that he answers at least 12 questions correctly
1
7630-7633
Suppose a student tosses a fair coin to determine his answer to each question If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false' Find the probability that he answers at least 12 questions correctly 8
1
7631-7634
If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false' Find the probability that he answers at least 12 questions correctly 8 Suppose X has a binomial distribution B 6, 21
1
7632-7635
Find the probability that he answers at least 12 questions correctly 8 Suppose X has a binomial distribution B 6, 21 Show that X = 3 is the most likely outcome
1
7633-7636
8 Suppose X has a binomial distribution B 6, 21 Show that X = 3 is the most likely outcome (Hint : P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6) 9
1
7634-7637
Suppose X has a binomial distribution B 6, 21 Show that X = 3 is the most likely outcome (Hint : P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6) 9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing
1
7635-7638
Show that X = 3 is the most likely outcome (Hint : P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6) 9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing 10
1
7636-7639
(Hint : P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6) 9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing 10 A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 100
1
7637-7640
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing 10 A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 100 What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice
1
7638-7641
10 A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 100 What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice Β© NCERT not to be republished 578 MATHEMATICS 11
1
7639-7642
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 100 What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice Β© NCERT not to be republished 578 MATHEMATICS 11 Find the probability of getting 5 exactly twice in 7 throws of a die
1
7640-7643
What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice Β© NCERT not to be republished 578 MATHEMATICS 11 Find the probability of getting 5 exactly twice in 7 throws of a die 12
1
7641-7644
Β© NCERT not to be republished 578 MATHEMATICS 11 Find the probability of getting 5 exactly twice in 7 throws of a die 12 Find the probability of throwing at most 2 sixes in 6 throws of a single die
1
7642-7645
Find the probability of getting 5 exactly twice in 7 throws of a die 12 Find the probability of throwing at most 2 sixes in 6 throws of a single die 13
1
7643-7646
12 Find the probability of throwing at most 2 sixes in 6 throws of a single die 13 It is known that 10% of certain articles manufactured are defective
1
7644-7647
Find the probability of throwing at most 2 sixes in 6 throws of a single die 13 It is known that 10% of certain articles manufactured are defective What is the probability that in a random sample of 12 such articles, 9 are defective
1
7645-7648
13 It is known that 10% of certain articles manufactured are defective What is the probability that in a random sample of 12 such articles, 9 are defective In each of the following, choose the correct answer: 14
1
7646-7649
It is known that 10% of certain articles manufactured are defective What is the probability that in a random sample of 12 such articles, 9 are defective In each of the following, choose the correct answer: 14 In a box containing 100 bulbs, 10 are defective
1
7647-7650
What is the probability that in a random sample of 12 such articles, 9 are defective In each of the following, choose the correct answer: 14 In a box containing 100 bulbs, 10 are defective The probability that out of a sample of 5 bulbs, none is defective is (A) 10–1 (B) βŽ›215 ⎞ ⎜ ⎟ ⎝ ⎠ (C) βŽ›1095 ⎞ ⎜ ⎟ ⎝ ⎠ (D) 9 10 15
1
7648-7651
In each of the following, choose the correct answer: 14 In a box containing 100 bulbs, 10 are defective The probability that out of a sample of 5 bulbs, none is defective is (A) 10–1 (B) βŽ›215 ⎞ ⎜ ⎟ ⎝ ⎠ (C) βŽ›1095 ⎞ ⎜ ⎟ ⎝ ⎠ (D) 9 10 15 The probability that a student is not a swimmer is 1
1
7649-7652
In a box containing 100 bulbs, 10 are defective The probability that out of a sample of 5 bulbs, none is defective is (A) 10–1 (B) βŽ›215 ⎞ ⎜ ⎟ ⎝ ⎠ (C) βŽ›1095 ⎞ ⎜ ⎟ ⎝ ⎠ (D) 9 10 15 The probability that a student is not a swimmer is 1 5 Then the probability that out of five students, four are swimmers is (A) 4 5 4 4 1 C 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (B) 44 1 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (C) 4 5 1 1 4 C 5 βŽ›5 ⎞ ⎜ ⎟ ⎝ ⎠ (D) None of these Miscellaneous Examples Example 33 Coloured balls are distributed in four boxes as shown in the following table: Box Colour Black White Red Blue I 3 4 5 6 II 2 2 2 2 III 1 2 3 1 IV 4 3 1 5 A box is selected at random and then a ball is randomly drawn from the selected box
1
7650-7653
The probability that out of a sample of 5 bulbs, none is defective is (A) 10–1 (B) βŽ›215 ⎞ ⎜ ⎟ ⎝ ⎠ (C) βŽ›1095 ⎞ ⎜ ⎟ ⎝ ⎠ (D) 9 10 15 The probability that a student is not a swimmer is 1 5 Then the probability that out of five students, four are swimmers is (A) 4 5 4 4 1 C 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (B) 44 1 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (C) 4 5 1 1 4 C 5 βŽ›5 ⎞ ⎜ ⎟ ⎝ ⎠ (D) None of these Miscellaneous Examples Example 33 Coloured balls are distributed in four boxes as shown in the following table: Box Colour Black White Red Blue I 3 4 5 6 II 2 2 2 2 III 1 2 3 1 IV 4 3 1 5 A box is selected at random and then a ball is randomly drawn from the selected box The colour of the ball is black, what is the probability that ball drawn is from the box III
1
7651-7654
The probability that a student is not a swimmer is 1 5 Then the probability that out of five students, four are swimmers is (A) 4 5 4 4 1 C 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (B) 44 1 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (C) 4 5 1 1 4 C 5 βŽ›5 ⎞ ⎜ ⎟ ⎝ ⎠ (D) None of these Miscellaneous Examples Example 33 Coloured balls are distributed in four boxes as shown in the following table: Box Colour Black White Red Blue I 3 4 5 6 II 2 2 2 2 III 1 2 3 1 IV 4 3 1 5 A box is selected at random and then a ball is randomly drawn from the selected box The colour of the ball is black, what is the probability that ball drawn is from the box III Β© NCERT not to be republished PROBABILITY 579 Solution Let A, E1, E2, E3 and E4 be the events as defined below : A : a black ball is selected E1 : box I is selected E2 : box II is selected E3 : box III is selected E4 : box IV is selected Since the boxes are chosen at random, Therefore P(E1) = P(E2) = P(E3) = P(E4) = 1 4 Also P(A|E1) = 3 18 , P(A|E2) = 2 8 , P(A|E3) = 1 7 and P(A|E4) = 4 13 P(box III is selected, given that the drawn ball is black) = P(E3|A)
1
7652-7655
5 Then the probability that out of five students, four are swimmers is (A) 4 5 4 4 1 C 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (B) 44 1 5 5 βŽ› ⎞ ⎜ ⎟ ⎝ ⎠ (C) 4 5 1 1 4 C 5 βŽ›5 ⎞ ⎜ ⎟ ⎝ ⎠ (D) None of these Miscellaneous Examples Example 33 Coloured balls are distributed in four boxes as shown in the following table: Box Colour Black White Red Blue I 3 4 5 6 II 2 2 2 2 III 1 2 3 1 IV 4 3 1 5 A box is selected at random and then a ball is randomly drawn from the selected box The colour of the ball is black, what is the probability that ball drawn is from the box III Β© NCERT not to be republished PROBABILITY 579 Solution Let A, E1, E2, E3 and E4 be the events as defined below : A : a black ball is selected E1 : box I is selected E2 : box II is selected E3 : box III is selected E4 : box IV is selected Since the boxes are chosen at random, Therefore P(E1) = P(E2) = P(E3) = P(E4) = 1 4 Also P(A|E1) = 3 18 , P(A|E2) = 2 8 , P(A|E3) = 1 7 and P(A|E4) = 4 13 P(box III is selected, given that the drawn ball is black) = P(E3|A) By Bayes' theorem, P(E3|A) = 3 3 1 1 2 2 3 3 4 4 P(E ) P(A|E ) P(E )P(A|E ) P(E )P(A|E )+P(E )P(A|E ) P(E )P(A|E ) = 1 1 4 7 0
1
7653-7656
The colour of the ball is black, what is the probability that ball drawn is from the box III Β© NCERT not to be republished PROBABILITY 579 Solution Let A, E1, E2, E3 and E4 be the events as defined below : A : a black ball is selected E1 : box I is selected E2 : box II is selected E3 : box III is selected E4 : box IV is selected Since the boxes are chosen at random, Therefore P(E1) = P(E2) = P(E3) = P(E4) = 1 4 Also P(A|E1) = 3 18 , P(A|E2) = 2 8 , P(A|E3) = 1 7 and P(A|E4) = 4 13 P(box III is selected, given that the drawn ball is black) = P(E3|A) By Bayes' theorem, P(E3|A) = 3 3 1 1 2 2 3 3 4 4 P(E ) P(A|E ) P(E )P(A|E ) P(E )P(A|E )+P(E )P(A|E ) P(E )P(A|E ) = 1 1 4 7 0 165 1 3 1 1 1 1 1 4 4 18 4 4 4 7 4 13 Γ— = Γ— + Γ— + Γ— + Γ— Example 34 Find the mean of the Binomial distribution B 4, 31
1
7654-7657
Β© NCERT not to be republished PROBABILITY 579 Solution Let A, E1, E2, E3 and E4 be the events as defined below : A : a black ball is selected E1 : box I is selected E2 : box II is selected E3 : box III is selected E4 : box IV is selected Since the boxes are chosen at random, Therefore P(E1) = P(E2) = P(E3) = P(E4) = 1 4 Also P(A|E1) = 3 18 , P(A|E2) = 2 8 , P(A|E3) = 1 7 and P(A|E4) = 4 13 P(box III is selected, given that the drawn ball is black) = P(E3|A) By Bayes' theorem, P(E3|A) = 3 3 1 1 2 2 3 3 4 4 P(E ) P(A|E ) P(E )P(A|E ) P(E )P(A|E )+P(E )P(A|E ) P(E )P(A|E ) = 1 1 4 7 0 165 1 3 1 1 1 1 1 4 4 18 4 4 4 7 4 13 Γ— = Γ— + Γ— + Γ— + Γ— Example 34 Find the mean of the Binomial distribution B 4, 31 Solution Let X be the random variable whose probability distribution is B 4,31
1
7655-7658
By Bayes' theorem, P(E3|A) = 3 3 1 1 2 2 3 3 4 4 P(E ) P(A|E ) P(E )P(A|E ) P(E )P(A|E )+P(E )P(A|E ) P(E )P(A|E ) = 1 1 4 7 0 165 1 3 1 1 1 1 1 4 4 18 4 4 4 7 4 13 Γ— = Γ— + Γ— + Γ— + Γ— Example 34 Find the mean of the Binomial distribution B 4, 31 Solution Let X be the random variable whose probability distribution is B 4,31 Here n = 4, p = 1 3 and q = 1 2 1 3 3 βˆ’ = We know that P(X = x) = 4 4 2 1 C 3 3 x x x βˆ’ βŽ› ⎞ βŽ› ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ , x = 0, 1, 2, 3, 4
1
7656-7659
165 1 3 1 1 1 1 1 4 4 18 4 4 4 7 4 13 Γ— = Γ— + Γ— + Γ— + Γ— Example 34 Find the mean of the Binomial distribution B 4, 31 Solution Let X be the random variable whose probability distribution is B 4,31 Here n = 4, p = 1 3 and q = 1 2 1 3 3 βˆ’ = We know that P(X = x) = 4 4 2 1 C 3 3 x x x βˆ’ βŽ› ⎞ βŽ› ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ , x = 0, 1, 2, 3, 4 i
1
7657-7660
Solution Let X be the random variable whose probability distribution is B 4,31 Here n = 4, p = 1 3 and q = 1 2 1 3 3 βˆ’ = We know that P(X = x) = 4 4 2 1 C 3 3 x x x βˆ’ βŽ› ⎞ βŽ› ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ , x = 0, 1, 2, 3, 4 i e
1
7658-7661
Here n = 4, p = 1 3 and q = 1 2 1 3 3 βˆ’ = We know that P(X = x) = 4 4 2 1 C 3 3 x x x βˆ’ βŽ› ⎞ βŽ› ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ , x = 0, 1, 2, 3, 4 i e the distribution of X is xi P(xi) xi P(xi) 0 4 4 0 2 C 3 0 1 3 4 1 2 1 C 3 3 3 4 1 2 1 C 3 3 Β© NCERT not to be republished 580 MATHEMATICS 2 2 2 4 2 2 1 C 3 3 2 2 4 2 2 1 2 C 3 3 3 3 4 3 2 1 C 3 3 3 4 3 2 1 3 C 3 3 4 4 4 4 1 C 3 4 4 4 1 4 C 3 Now Mean (ΞΌ) = 4 1 ( ) i i i x p x =βˆ‘ = 3 2 2 4 4 1 2 2 1 2 1 0 C 2 C 3 3 3 3 βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ + + β‹… ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + 3 4 4 4 3 4 2 1 1 3 C 4 C 3 3 3 = 3 2 4 4 4 4 2 2 2 1 4 2 6 3 4 4 1 3 3 3 3 Γ— + Γ— Γ— + Γ— Γ— + Γ— Γ— = 4 32 48 24 4 108 4 81 3 3 + + + = = Example 35 The probability of a shooter hitting a target is 3 4
1
7659-7662
i e the distribution of X is xi P(xi) xi P(xi) 0 4 4 0 2 C 3 0 1 3 4 1 2 1 C 3 3 3 4 1 2 1 C 3 3 Β© NCERT not to be republished 580 MATHEMATICS 2 2 2 4 2 2 1 C 3 3 2 2 4 2 2 1 2 C 3 3 3 3 4 3 2 1 C 3 3 3 4 3 2 1 3 C 3 3 4 4 4 4 1 C 3 4 4 4 1 4 C 3 Now Mean (ΞΌ) = 4 1 ( ) i i i x p x =βˆ‘ = 3 2 2 4 4 1 2 2 1 2 1 0 C 2 C 3 3 3 3 βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ + + β‹… ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + 3 4 4 4 3 4 2 1 1 3 C 4 C 3 3 3 = 3 2 4 4 4 4 2 2 2 1 4 2 6 3 4 4 1 3 3 3 3 Γ— + Γ— Γ— + Γ— Γ— + Γ— Γ— = 4 32 48 24 4 108 4 81 3 3 + + + = = Example 35 The probability of a shooter hitting a target is 3 4 How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0
1
7660-7663
e the distribution of X is xi P(xi) xi P(xi) 0 4 4 0 2 C 3 0 1 3 4 1 2 1 C 3 3 3 4 1 2 1 C 3 3 Β© NCERT not to be republished 580 MATHEMATICS 2 2 2 4 2 2 1 C 3 3 2 2 4 2 2 1 2 C 3 3 3 3 4 3 2 1 C 3 3 3 4 3 2 1 3 C 3 3 4 4 4 4 1 C 3 4 4 4 1 4 C 3 Now Mean (ΞΌ) = 4 1 ( ) i i i x p x =βˆ‘ = 3 2 2 4 4 1 2 2 1 2 1 0 C 2 C 3 3 3 3 βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ + + β‹… ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + 3 4 4 4 3 4 2 1 1 3 C 4 C 3 3 3 = 3 2 4 4 4 4 2 2 2 1 4 2 6 3 4 4 1 3 3 3 3 Γ— + Γ— Γ— + Γ— Γ— + Γ— Γ— = 4 32 48 24 4 108 4 81 3 3 + + + = = Example 35 The probability of a shooter hitting a target is 3 4 How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0 99
1
7661-7664
the distribution of X is xi P(xi) xi P(xi) 0 4 4 0 2 C 3 0 1 3 4 1 2 1 C 3 3 3 4 1 2 1 C 3 3 Β© NCERT not to be republished 580 MATHEMATICS 2 2 2 4 2 2 1 C 3 3 2 2 4 2 2 1 2 C 3 3 3 3 4 3 2 1 C 3 3 3 4 3 2 1 3 C 3 3 4 4 4 4 1 C 3 4 4 4 1 4 C 3 Now Mean (ΞΌ) = 4 1 ( ) i i i x p x =βˆ‘ = 3 2 2 4 4 1 2 2 1 2 1 0 C 2 C 3 3 3 3 βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ βŽ› ⎞ + + β‹… ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + 3 4 4 4 3 4 2 1 1 3 C 4 C 3 3 3 = 3 2 4 4 4 4 2 2 2 1 4 2 6 3 4 4 1 3 3 3 3 Γ— + Γ— Γ— + Γ— Γ— + Γ— Γ— = 4 32 48 24 4 108 4 81 3 3 + + + = = Example 35 The probability of a shooter hitting a target is 3 4 How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0 99 Solution Let the shooter fire n times
1
7662-7665
How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0 99 Solution Let the shooter fire n times Obviously, n fires are n Bernoulli trials
1
7663-7666
99 Solution Let the shooter fire n times Obviously, n fires are n Bernoulli trials In each trial, p = probability of hitting the target = 3 4 and q = probability of not hitting the target = 1 4
1
7664-7667
Solution Let the shooter fire n times Obviously, n fires are n Bernoulli trials In each trial, p = probability of hitting the target = 3 4 and q = probability of not hitting the target = 1 4 Then P(X = x) = 1 3 3 C C C 4 4 4 n x x x n n x x n n x x x n q p βˆ’ βˆ’ βŽ› ⎞ βŽ› ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
1
7665-7668
Obviously, n fires are n Bernoulli trials In each trial, p = probability of hitting the target = 3 4 and q = probability of not hitting the target = 1 4 Then P(X = x) = 1 3 3 C C C 4 4 4 n x x x n n x x n n x x x n q p βˆ’ βˆ’ βŽ› ⎞ βŽ› ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Now, given that, P(hitting the target at least once) > 0
1
7666-7669
In each trial, p = probability of hitting the target = 3 4 and q = probability of not hitting the target = 1 4 Then P(X = x) = 1 3 3 C C C 4 4 4 n x x x n n x x n n x x x n q p βˆ’ βˆ’ βŽ› ⎞ βŽ› ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Now, given that, P(hitting the target at least once) > 0 99 i
1
7667-7670
Then P(X = x) = 1 3 3 C C C 4 4 4 n x x x n n x x n n x x x n q p βˆ’ βˆ’ βŽ› ⎞ βŽ› ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Now, given that, P(hitting the target at least once) > 0 99 i e
1
7668-7671
Now, given that, P(hitting the target at least once) > 0 99 i e P(x β‰₯ 1) > 0
1
7669-7672
99 i e P(x β‰₯ 1) > 0 99 Β© NCERT not to be republished PROBABILITY 581 Therefore, 1 – P (x = 0) > 0
1
7670-7673
e P(x β‰₯ 1) > 0 99 Β© NCERT not to be republished PROBABILITY 581 Therefore, 1 – P (x = 0) > 0 99 or 0 1 1 C 4 n n βˆ’ > 0
1
7671-7674
P(x β‰₯ 1) > 0 99 Β© NCERT not to be republished PROBABILITY 581 Therefore, 1 – P (x = 0) > 0 99 or 0 1 1 C 4 n n βˆ’ > 0 99 or 0 1 1 C 0
1
7672-7675
99 Β© NCERT not to be republished PROBABILITY 581 Therefore, 1 – P (x = 0) > 0 99 or 0 1 1 C 4 n n βˆ’ > 0 99 or 0 1 1 C 0 01 i
1
7673-7676
99 or 0 1 1 C 4 n n βˆ’ > 0 99 or 0 1 1 C 0 01 i e
1
7674-7677
99 or 0 1 1 C 0 01 i e 4 4 n n n < 0
1
7675-7678
01 i e 4 4 n n n < 0 01 or 4n > 1 0
1
7676-7679
e 4 4 n n n < 0 01 or 4n > 1 0 01 = 100
1
7677-7680
4 4 n n n < 0 01 or 4n > 1 0 01 = 100 (1) The minimum value of n to satisfy the inequality (1) is 4
1
7678-7681
01 or 4n > 1 0 01 = 100 (1) The minimum value of n to satisfy the inequality (1) is 4 Thus, the shooter must fire 4 times
1
7679-7682
01 = 100 (1) The minimum value of n to satisfy the inequality (1) is 4 Thus, the shooter must fire 4 times Example 36 A and B throw a die alternatively till one of them gets a β€˜6’ and wins the game
1
7680-7683
(1) The minimum value of n to satisfy the inequality (1) is 4 Thus, the shooter must fire 4 times Example 36 A and B throw a die alternatively till one of them gets a β€˜6’ and wins the game Find their respective probabilities of winning, if A starts first
1
7681-7684
Thus, the shooter must fire 4 times Example 36 A and B throw a die alternatively till one of them gets a β€˜6’ and wins the game Find their respective probabilities of winning, if A starts first Solution Let S denote the success (getting a β€˜6’) and F denote the failure (not getting a β€˜6’)
1
7682-7685
Example 36 A and B throw a die alternatively till one of them gets a β€˜6’ and wins the game Find their respective probabilities of winning, if A starts first Solution Let S denote the success (getting a β€˜6’) and F denote the failure (not getting a β€˜6’) Thus, P(S) = 1 5 6, P(F) 6 = P(A wins in the first throw) = P(S) = 1 6 A gets the third throw, when the first throw by A and second throw by B result into failures
1
7683-7686
Find their respective probabilities of winning, if A starts first Solution Let S denote the success (getting a β€˜6’) and F denote the failure (not getting a β€˜6’) Thus, P(S) = 1 5 6, P(F) 6 = P(A wins in the first throw) = P(S) = 1 6 A gets the third throw, when the first throw by A and second throw by B result into failures Therefore, P(A wins in the 3rd throw) = P(FFS) = 5 5 1 P(F)P(F)P(S)= 6 6 6 = 52 1 6 6 βŽ› ⎞ Γ— ⎜ ⎟ ⎝ ⎠ P(A wins in the 5th throw) = P (FFFFS) 54 1 6 6 and so on
1
7684-7687
Solution Let S denote the success (getting a β€˜6’) and F denote the failure (not getting a β€˜6’) Thus, P(S) = 1 5 6, P(F) 6 = P(A wins in the first throw) = P(S) = 1 6 A gets the third throw, when the first throw by A and second throw by B result into failures Therefore, P(A wins in the 3rd throw) = P(FFS) = 5 5 1 P(F)P(F)P(S)= 6 6 6 = 52 1 6 6 βŽ› ⎞ Γ— ⎜ ⎟ ⎝ ⎠ P(A wins in the 5th throw) = P (FFFFS) 54 1 6 6 and so on Hence, P(A wins) = 2 4 1 5 1 5 1
1
7685-7688
Thus, P(S) = 1 5 6, P(F) 6 = P(A wins in the first throw) = P(S) = 1 6 A gets the third throw, when the first throw by A and second throw by B result into failures Therefore, P(A wins in the 3rd throw) = P(FFS) = 5 5 1 P(F)P(F)P(S)= 6 6 6 = 52 1 6 6 βŽ› ⎞ Γ— ⎜ ⎟ ⎝ ⎠ P(A wins in the 5th throw) = P (FFFFS) 54 1 6 6 and so on Hence, P(A wins) = 2 4 1 5 1 5 1 6 6 6 6 6 = 61 25 1 36 βˆ’ = 6 11 Β© NCERT not to be republished 582 MATHEMATICS P(B wins) = 1 – P (A wins) = 6 1 11 115 Remark If a + ar + ar2 +
1
7686-7689
Therefore, P(A wins in the 3rd throw) = P(FFS) = 5 5 1 P(F)P(F)P(S)= 6 6 6 = 52 1 6 6 βŽ› ⎞ Γ— ⎜ ⎟ ⎝ ⎠ P(A wins in the 5th throw) = P (FFFFS) 54 1 6 6 and so on Hence, P(A wins) = 2 4 1 5 1 5 1 6 6 6 6 6 = 61 25 1 36 βˆ’ = 6 11 Β© NCERT not to be republished 582 MATHEMATICS P(B wins) = 1 – P (A wins) = 6 1 11 115 Remark If a + ar + ar2 + + arn–1 +
1
7687-7690
Hence, P(A wins) = 2 4 1 5 1 5 1 6 6 6 6 6 = 61 25 1 36 βˆ’ = 6 11 Β© NCERT not to be republished 582 MATHEMATICS P(B wins) = 1 – P (A wins) = 6 1 11 115 Remark If a + ar + ar2 + + arn–1 + , where |r| < 1, then sum of this infinite G
1
7688-7691
6 6 6 6 6 = 61 25 1 36 βˆ’ = 6 11 Β© NCERT not to be republished 582 MATHEMATICS P(B wins) = 1 – P (A wins) = 6 1 11 115 Remark If a + ar + ar2 + + arn–1 + , where |r| < 1, then sum of this infinite G P
1
7689-7692
+ arn–1 + , where |r| < 1, then sum of this infinite G P is given by
1
7690-7693
, where |r| < 1, then sum of this infinite G P is given by 1 a βˆ’r (Refer A
1
7691-7694
P is given by 1 a βˆ’r (Refer A 1
1
7692-7695
is given by 1 a βˆ’r (Refer A 1 3 of Class XI Text book)
1
7693-7696
1 a βˆ’r (Refer A 1 3 of Class XI Text book) Example 37 If a machine is correctly set up, it produces 90% acceptable items
1
7694-7697
1 3 of Class XI Text book) Example 37 If a machine is correctly set up, it produces 90% acceptable items If it is incorrectly set up, it produces only 40% acceptable items
1
7695-7698
3 of Class XI Text book) Example 37 If a machine is correctly set up, it produces 90% acceptable items If it is incorrectly set up, it produces only 40% acceptable items Past experience shows that 80% of the set ups are correctly done
1
7696-7699
Example 37 If a machine is correctly set up, it produces 90% acceptable items If it is incorrectly set up, it produces only 40% acceptable items Past experience shows that 80% of the set ups are correctly done If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup
1
7697-7700
If it is incorrectly set up, it produces only 40% acceptable items Past experience shows that 80% of the set ups are correctly done If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup Solution Let A be the event that the machine produces 2 acceptable items
1
7698-7701
Past experience shows that 80% of the set ups are correctly done If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup Solution Let A be the event that the machine produces 2 acceptable items Also let B1 represent the event of correct set up and B2 represent the event of incorrect setup
1
7699-7702
If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup Solution Let A be the event that the machine produces 2 acceptable items Also let B1 represent the event of correct set up and B2 represent the event of incorrect setup Now P(B1) = 0
1
7700-7703
Solution Let A be the event that the machine produces 2 acceptable items Also let B1 represent the event of correct set up and B2 represent the event of incorrect setup Now P(B1) = 0 8, P(B2) = 0
1
7701-7704
Also let B1 represent the event of correct set up and B2 represent the event of incorrect setup Now P(B1) = 0 8, P(B2) = 0 2 P(A|B1) = 0
1
7702-7705
Now P(B1) = 0 8, P(B2) = 0 2 P(A|B1) = 0 9 Γ— 0
1
7703-7706
8, P(B2) = 0 2 P(A|B1) = 0 9 Γ— 0 9 and P(A|B2) = 0
1
7704-7707
2 P(A|B1) = 0 9 Γ— 0 9 and P(A|B2) = 0 4 Γ— 0
1
7705-7708
9 Γ— 0 9 and P(A|B2) = 0 4 Γ— 0 4 Therefore P(B1|A) = 1 1 1 1 2 2 P(B ) P(A|B ) P(B ) P(A|B ) + P(B ) P(A|B ) = 0
1
7706-7709
9 and P(A|B2) = 0 4 Γ— 0 4 Therefore P(B1|A) = 1 1 1 1 2 2 P(B ) P(A|B ) P(B ) P(A|B ) + P(B ) P(A|B ) = 0 8Γ— 0
1
7707-7710
4 Γ— 0 4 Therefore P(B1|A) = 1 1 1 1 2 2 P(B ) P(A|B ) P(B ) P(A|B ) + P(B ) P(A|B ) = 0 8Γ— 0 9Γ— 0
1
7708-7711
4 Therefore P(B1|A) = 1 1 1 1 2 2 P(B ) P(A|B ) P(B ) P(A|B ) + P(B ) P(A|B ) = 0 8Γ— 0 9Γ— 0 9 648 0
1
7709-7712
8Γ— 0 9Γ— 0 9 648 0 95 0
1
7710-7713
9Γ— 0 9 648 0 95 0 8Γ— 0
1
7711-7714
9 648 0 95 0 8Γ— 0 9Γ— 0
1
7712-7715
95 0 8Γ— 0 9Γ— 0 9 + 0
1
7713-7716
8Γ— 0 9Γ— 0 9 + 0 2Γ— 0
1
7714-7717
9Γ— 0 9 + 0 2Γ— 0 4Γ— 0
1
7715-7718
9 + 0 2Γ— 0 4Γ— 0 4 =680 = Miscellaneous Exercise on Chapter 13 1
1
7716-7719
2Γ— 0 4Γ— 0 4 =680 = Miscellaneous Exercise on Chapter 13 1 A and B are two events such that P (A) β‰  0
1
7717-7720
4Γ— 0 4 =680 = Miscellaneous Exercise on Chapter 13 1 A and B are two events such that P (A) β‰  0 Find P(B|A), if (i) A is a subset of B (ii) A ∩ B = Ο† 2