Question

A coach is training 3 players. He observes that the player A can hit a target 4 times in 5 shots, player B can hit 3 times in 4 shots and the player C can hit 2 times in 3 shots.

From this situation answer the following:

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Question 1

Let the target is hit by A, B and C. Then, the probability that A, B and, C all will hit, is

(a) 4/5 

(b) 3/5 

(c) 2/5 

(d) 1/5

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Question 2

Referring to (i), what is the probability that B, C will hit and A will lose?

(a) 1/10 

(b) 3/10 

(c) 7/10 

(d) 4/10

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Question 3

With reference to the events mentioned in (i), what is the probability that ‘any two of A, B and C will hit?

(a) 1/30 

(b) 11/30 

(c) 17/30 

(d) 13/30

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Question 4

What is the probability that ‘none of them will hit the target’?

(a) 1/30 

(b) 1/60 

(c) 1/15 

(d) 2/15

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Question 5

What is the probability that at least one of A, B or C will hit the target?

(a) 59/60 

(b) 2/5 

(c) 3/5 

(d) 1/60

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Transcript

Question A coach is training 3 players. He observes that the player A can hit a target 4 times in 5 shots, player B can hit 3 times in 4 shots and the player C can hit 2 times in 3 shots. From this situation answer the following:P(A) = Probability of A hitting the target = 𝟒/𝟓 P(B) = Probability of B hitting the target = 𝟑/𝟒 P(C) = Probability of C hitting the target = 𝟐/𝟑 Question 1 Let the target is hit by A, B and C. Then, the probability that A, B and, C all will hit, is (a) 4/5 (b) 3/5 (c) 2/5 (d) 1/5P(all hit) = Probability A will hit × Probability B will hit × Probability C will hit = 4/5 × 3/4 × 2/3 = 𝟐/𝟓 So, the correct answer is (c) Question 2 Referring to (i), what is the probability that B, C will hit and A will lose? (a) 1/10 (b) 3/10 (c) 7/10 (d) 4/10P(B and C will hit and A will lose) = Probability A will not hit × Probability B will hit × Probability C will hit = (𝟏−𝟒/𝟓)"×" 𝟑/𝟒 × 𝟐/𝟑 = 1/5 × 3/4 × 2/3 = 𝟏/𝟏𝟎 So, the correct answer is (a) Question 3 With reference to the events mentioned in (i), what is the probability that ‘any two of A, B and C will hit? (a) 1/30 (b) 11/30 (c) 17/30 (d) 13/30 P(any two will hit) = P(A will not hit) × P(B will hit) × P(C will hit) + P(A will hit) × P(B will not hit) × P(C will hit) + P(A will hit) × P(B will hit) × P(C will not hit) = (1−4/5)"×" 3/4 × 2/3 + 4/5 "×" (1−3/4)× 2/3 +4/5 "×" 3/4 × (1−2/3) = 1/5 "×" 3/4 × 2/3 + 4/5 "×" 1/4 × 2/3 +4/5 "×" 3/4 × 1/3 = 1/10+2/15+1/5 = 𝟏𝟑/𝟑𝟎 So, the correct answer is (d) Question 4 What is the probability that ‘none of them will hit the target’? (a) 1/30 (b) 1/60 (c) 1/15 (d) 2/15P(none will hit) = Probability A will not hit × Probability B will not hit × Probability C will not hit = (𝟏−𝟒/𝟓) × (𝟏−𝟑/𝟒)×(𝟏−𝟐/𝟑) = 1/5×1/4×1/3 = 𝟏/𝟔𝟎 So, the correct answer is (b) Question 5 What is the probability that at least one of A, B or C will hit the target? (a) 59/60 (b) 2/5 (c) 3/5 (d) 1/60P( at least one will hit) = P(exactly one will hit) + P(exactly two will hit) + P(all will hit) P(exactly one will hit) P(exactly one will hit) = P(A will hit)×P(B will not hit) × P(C will not hit) + P (A will not hit) × P (B will hit) × P (C will not hit) + P (A will not hit)× P (B will not hit) × P (C will hit) = 4/5×(1−3/4)×(1−2/3) +(1−4/5)×3/4×(1−2/3) +(1−4/5)×(1−3/4)×2/3= 4/5 "×" 1/4 × 1/3 + 1/5 "×" 3/4× 1/3 +1/5 "×" 1/4 × 2/3 = 1/15+1/20+1/30 = 𝟗/𝟔𝟎P(exactly two will hit) This is calculated in Question 3 P(exactly two will hit) = 𝟏𝟑/𝟑𝟎 P(all will hit) This is calculated in Question 1 P(all will hit) = 𝟐/𝟓Now, P(at least one will hit) = P(exactly one will hit) + P(exactly two will hit) + P(all will hit) = 9/60+13/30+2/5 = 𝟓𝟗/𝟔𝟎 So, the correct answer is (a)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.