Vtyjkyuk

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?

HINT:

$textP(Required)= P(A and B hits, C misses)+P(A and C hits, B misses) +$

$;;;;;;;;;;;;;;;;;;;;;;textP(C and B hits, A misses)+P(A,B,C all hit)$

$textP(A)=A hits
, P(A')=A misses= 1- P(A)$

$textP(B)=B hits
, P(B')=B misses= 1- P(B)$

$textP(C)=C hits
, P(C')=C misses= 1- P(C)$

So you can write,

$textP(Required)= P(A)P(B)P(C')+P(A)P(B')P(C) +P(A')P(B)P(C)+P(A)P(B)P(C)$

$$=frac12cdotfrac13cdotbigg(1-frac14bigg)+frac12cdotfrac14cdotbigg(1-frac13bigg)+frac14cdotfrac13cdotbigg(1-frac12bigg)+frac12cdotfrac13cdotfrac14$$

$$=frac324+frac224+frac124+frac124$$

$$=frac724$$

Comment if you have any more doubts.

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.
$$P(mathrmA miss, B miss, C miss)=P(mathrmA miss) cdot P(mathrmB hit) cdot P(mathrmC miss) = (1-frac12) cdot frac13 cdot (1-frac14)$$

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 :)