Probability 3

Probability Level 4

Three shots are fired independently at the target in succession.The probabilities that the target is hit in first shot is $\frac{1}{2}$ ,in the second $\frac{2}{3}$ and in third shot is $\frac{3}{4}$ .In case of exactly one hit ,the probability of destroying the target is $\frac{1}{3}$ and in case of exactly two hits, $\frac{7}{11}$ and in the case of three hits is $1.0$ .Find the probability of destroying in three shots.

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The answer is 0.625.

1 solution

Ishan Tandon
Nov 6, 2020

I feel like the answer is NOT correct.

Your Solution:

" ' " Signifies a HIT

E1 : Destroyed in 1 Hit

1 Hit : A'BCE1 + AB'CE1 + ABC'E1

Problem : You are assuming that even thought the target is destroyed, subsequent hits occur

Ex: A'BCE1 -> A hit destroys the target but even then B and C hit occur after that in order, Even though the target they wanna hit is gone

Correction : Once a target is destroyed furthur hits dont occur => For the one hit case one C' is viable otherwise 3 Hits won't occur

Assuming my argument is correct answer would be : $\frac{23}{88}$ ~ 0.261

Probability 3

The answer is 0.625.

1 solution

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