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Three persons A, B and C aim at a target. The probability of person A, B, and C hitting the target is $1/2$, $1/3$, and $1/4$ respectively. Find the probability that at least two of them will hit the target?

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    $\begingroup$ Hello and welcome to Mathematics SE! Can you show us what you have tried so far? And what is your knowledge of this subject? Can you imagine how you would find the probability that they all hit or miss? $\endgroup$
    – Jan
    Aug 21, 2018 at 14:22
  • $\begingroup$ Can you find the probability that A and B hit and C misses? $\endgroup$ Aug 21, 2018 at 14:28
  • $\begingroup$ Aayush. please upvote and accept (green tick) an answer if your query has been answered. $\endgroup$
    – prog_SAHIL
    Aug 22, 2018 at 3:02

2 Answers 2

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HINT:

$\text{P(Required)= P(A and B hits, C misses)+P(A and C hits, B misses) +}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{P(C and B hits, A misses)+P(A,B,C all hit)}$

$\text{P(A)=A hits , P(A')=A misses= 1- P(A)}$

$\text{P(B)=B hits , P(B')=B misses= 1- P(B)}$

$\text{P(C)=C hits , P(C')=C misses= 1- P(C)}$

So you can write,

$\text{P(Required)= P(A)P(B)P(C')+P(A)P(B')P(C) +P(A')P(B)P(C)+P(A)P(B)P(C)}$

$$=\frac{1}{2}\cdot\frac{1}{3}\cdot\bigg(1-\frac{1}{4}\bigg)+\frac{1}{2}\cdot\frac{1}{4}\cdot\bigg(1-\frac{1}{3}\bigg)+\frac{1}{4}\cdot\frac{1}{3}\cdot\bigg(1-\frac{1}{2}\bigg)+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}$$

$$=\frac{3}{24}+\frac{2}{24}+\frac{1}{24}+\frac{1}{24}$$

$$=\frac{7}{24}$$

Comment if you have any more doubts.

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  • $\begingroup$ Answer is 7/24 but I don't know how it came? I wants steps of solution it would be great fun for me if can solve it with steps of solution. $\endgroup$ Aug 21, 2018 at 15:19
  • $\begingroup$ @AayushMukharji are you familiar with independent and dependent events? $\endgroup$
    – prog_SAHIL
    Aug 21, 2018 at 15:22
  • $\begingroup$ Good explanation. I added my answer as a more basic version to guide the OP who does not appear to have a solid background in probability and try to give a more intuitive approach. $\endgroup$
    – Jan
    Aug 21, 2018 at 15:29
  • $\begingroup$ @Jan, He is most probably unaware of how independent and dependent events work. Nice answer. +1'ed. $\endgroup$
    – prog_SAHIL
    Aug 21, 2018 at 15:31
  • $\begingroup$ Thank you I understood everything.Thanks for your quick reply.This is the best website and best people here answering and discussing. $\endgroup$ Aug 22, 2018 at 1:50
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I am not going to be giving you the answer, but I will hopefully give you the means to find it yourself.

So we have three shooters who can all hit or miss the target. So there are many possible outcomes of the shooting. For example, A can miss, B can hit and C can miss. Clearly this is different from A hitting, B missing and C missing. Even though the amount of them hitting the target in total is the same, the events are not the same. In fact, they don't even have the same probability!

The first event, A miss, B hit, C miss has a certain probability. Do you know how to compute this? Assuming that they shoot independently (A hitting or missing does not impact the chances of B or C hitting or missing) we can multiply the probabilities to get the probability of this event. $$P(\mathrm{A \ miss, \ B \ miss, \ C \ miss})=P(\mathrm{A\ miss}) \cdot P(\mathrm{B \ hit}) \cdot P(\mathrm{C\ miss}) = (1-\frac{1}{2}) \cdot \frac{1}{3} \cdot (1-\frac{1}{4})$$

Now see if you can compute this for the other event.

Now that you understand how to calculate the probability for an event, let's look at the question.

We are looking for the probability of 2 or more shooters hitting the target. So this is the cases where 2 of the 3 shooters hit and the case where they all hit the target. If you list all these events, you see that there are 4 events in total for which you will have to compute the probabilities. Adding them together will give you the required answer!

Good luck :)

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  • $\begingroup$ Thanks for guiding me. $\endgroup$ Aug 22, 2018 at 1:48

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