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Three shooters shoot at the same target, each of them shoots just once. The first one hits the target with a probability of $70\%$, the second one with a probability of $80\%$, and the third one with a probability of $90\%$. What is the probability that the shooters will hit the target

a) at least once?

b) at least twice?

I don't understand well this problem. If each shooter shoots at the target once, how can each of them shoot at it twice?

Do you know how to solve this problem?

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  • $\begingroup$ You must combine the results of the three shooters into a single score and calculate the probabilities. $\endgroup$
    – M. Wind
    Nov 3, 2018 at 20:14
  • $\begingroup$ b) asks about the probability of at least two of the shooters hitting the target $\endgroup$
    – Patricio
    Nov 3, 2018 at 20:23
  • $\begingroup$ The question is asking for (a) the probability that at least one of the three shooters hits the target, and (b) the probability that at least two of the three shooters hit the target. $\endgroup$
    – Brian Tung
    Nov 3, 2018 at 20:30
  • $\begingroup$ Does anyone knows how to solve part b of this problem? $\endgroup$
    – James Warthington
    Nov 3, 2018 at 21:24

3 Answers 3

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Your solution to the first problem is correct.

Let $A$ be the event that the first shooter hits the target; let $B$ be the event that the second shooter hits the target; let $C$ be the event that the third shooter hits the target. If at least two shooters hit the target, then either exactly two of them hit the target or all three do. Thus, we must calculate $$\Pr(A)\Pr(B)\Pr(C^C) + \Pr(A)\Pr(B^C)\Pr(C) + \Pr(A^C)\Pr(B)\Pr(C) + \Pr(A)\Pr(B)\Pr(C)$$ given $\Pr(A) = 0.70$, $\Pr(B) = 0.80$, $\Pr(C) = 0.90$. Can you proceed?

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    $\begingroup$ Thank you, my original line of reasoning is finally rescued by you. Thanks! $\endgroup$
    – James Warthington
    Nov 3, 2018 at 22:49
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For part 2

Probability $$~(p(s1)*p(s2)*(1-p(s3))) +((1-p(s1))*p(s2)*p(s3)) +(p(s1)*(1-p(s2))*p(s3)) + (p(s1)*p(s2)*(p(s3)))~$$

so $$(0.7*0.8*0.1)+(0.7*0.2*0.9)+(0.3*0.8*0.9)+(0.9*0.8*0.7) = 0.902 $$

90%

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I figure out how to do a) The probability of 3 shooters at least hitting target once is equal to 1 - probability that no one hits the target. So it is $1-(0.2*0.3*0.1)=0.994$

How do you solve b)

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  • $\begingroup$ Since you know how to solve part (a), I suggest you include this solution in your question, explain that you are trying to solve part (b), and delete the second question you asked about solving part (b). $\endgroup$
    – N. F. Taussig
    Nov 3, 2018 at 21:42
  • $\begingroup$ Can you give me a hint to part b $\endgroup$
    – James Warthington
    Nov 3, 2018 at 22:06
  • $\begingroup$ For part b, my way of thinking is actually failed. The probability that shooters hit target twice is equal the probability that shooter 1 and shooter 2 hits the target plus the probability that shooter 2 and shooter 3 hits the target plus the probability that shooter 1 and shooter 3 hits the target. The sum of these 3 probabilities is greater than 1, so my line of reasoning fails miserably here. $\endgroup$
    – James Warthington
    Nov 3, 2018 at 22:13

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