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Three shooters shoot a target. The known probability to hit the target of the first, second and third shooter are 0.3, 0.4 and 0.5 respectively. What is the probability that the target is hit from only one shooter?

I try this.

The probabilities that two will not heat the target:

0.7 x 0.6 = 0.42 0.7 x 0.5 = 0.35 0.6 x 0.5 = 0.30

and finally I'm not sure what should i do with those results? suppose have to multiply them?

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  • $\begingroup$ Can you calculate the probability that the first shooter hit the target and the other don't? You should easily arrive to the result. $\endgroup$
    – nicola
    Dec 13, 2016 at 15:44
  • $\begingroup$ How many ways there are that only 1 of them hit it? What are the probabilities for each of those outcomes? Multiply the probability of one of them hitting and the other 2 missing for each case. $\endgroup$
    – Jesús Ros
    Dec 13, 2016 at 15:45

2 Answers 2

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Split it into disjoint events, and then add up their probabilities:

  • The probability that the target is hit by the 1st shooter only is $\small(0.3)\times(1-0.4)\times(1-0.5)$
  • The probability that the target is hit by the 2nd shooter only is $\small(1-0.3)\times(0.4)\times(1-0.5)$
  • The probability that the target is hit by the 3rd shooter only is $\small(1-0.3)\times(1-0.4)\times(0.5)$

The overall probability is therefore:

$$\small(0.3)\times(1-0.4)\times(1-0.5)+(1-0.3)\times(0.4)\times(1-0.5)+(1-0.3)\times(1-0.4)\times(0.5)=0.44$$

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Let 3 shoots A, B and C.

$P(A) = 0.3, P(A') = 1 - 0.3 = 0.7$

$P(B) = 0.4, P(B') = 1 - 0.4 = 0.6$

$P(C) = 0.5, P(C') = 1 - 0.5 = 0.5$

Probability of hitting by only one = $P(A) \times P(B') \times P(C') + P(B) \times P(C') \times P(A') + P(C) \times P(A') \times P(B')$

$= 0.3 \times 0.6 \times 0.5 + 0.4 \times 0.5 \times 0.7 + 0.5 \times 0.7 \times 0.6$

$= 0.09 + 0.14 + 0.21 = 0.44$

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  • $\begingroup$ thank you for the detailed answer $\endgroup$
    – vachoh
    Dec 13, 2016 at 20:32
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    $\begingroup$ Mine pleasure. And I saw your questions. You not accept any answer in your all questions. I think you have to accept the answers (use tick). Because in future if anyone have similar problem. They get help from your questions easily. $\endgroup$ Dec 14, 2016 at 2:38
  • $\begingroup$ @KanwaljitSingh why can't we use inclusion-exclusion here? $P(A) + P(B) + P(C) - P(A,B) - P(A,C) - P(B,C) + P(A,B,C)$ ? $\endgroup$
    – Mining
    Sep 23, 2021 at 9:30

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