A,B & C shoot to hit a target. If A hits the target 4times in 5 trials; B hits it 3 times in 4 trials and C hits it 2 times in 3 trials, what is the probability that the target is hit by at least 2 persons? Explain also.
Answers
Answer:
5/6
Step-by-step explanation:
A hits the target 4 times in 5 trials.
P(A) = 4/5
P(A') = 1 - 4/5 = 1/5
B hits target in 3 times in 4 trials.
P(B) = 3/5
P(B') = 1 - 3/5 = 2/5
C hits target in 2 times in 3 trials.
P(C) = 2/3.
P(C') = 1 - 2/3 = 1/3
We have to find probability that target is hit by atleast 2 people. For that we should solve remaining cases also.
case 1.
Probability that A,B,C hit target : P(A∩B∩C)
=> P(A)P(B)(P(C)
=> 4/5 * 3/4 * 2/3
=> 2/5
Case 2.
Probability that A and B hit the target but not C : P(A∩B∩C')
=> P(A)P(B)P(C')
=> 4/5 * 3/4 * 1/3
=> 1/5
Case 3.
Probability that A and C hit target but not B : P(A∩C∩B')
=> P(A)P(C)P(B')
=> 4/5 * 2/3 * 1/4
=> 2/15
Case 4.
Probability that B and C hit target but not A : P(B∩C∩A')
=> P(B)P(C)P(A')
=> 3/5 * 2/3 * 1/4
=> 1/10
We now calculate the probability of target hit by atleast 2 people.
=> 2/5 + 1/5 + 2/15 + 1/10
=> (12+6+4+3)/30
=> 25/30
=> 5/6
Hence, required probability = 5/6
#Hope my answer helped you.
Step-by-step explanation:
A hits the target 4 times in 5 trials.
P(A) = 4/5
P(A') = 1 - 4/5 = 1/5
B hits target in 3 times in 4 trials.
P(B) = 3/5
P(B') = 1 - 3/5 = 2/5
C hits target in 2 times in 3 trials.
P(C) = 2/3.
P(C') = 1 - 2/3 = 1/3
We have to find probability that target is hit by atleast 2 people. For that we should solve remaining cases also.
case 1.
Probability that A,B,C hit target : P(A∩B∩C)
=> P(A)P(B)(P(C)
=> 4/5 * 3/4 * 2/3
=> 2/5