Latest Elitmus Aptitude Question SOLUTION: A man can hit the target once in four shots. If he fires four shots in succession, what is the probability that he will hit the target? a) 1 b) 1/256 c) 81/256 d) 175/256

In four shots he can hit once,twice,thrice,all hit
the probability of hitting the target isp(1hit out of 4)+P(2hit out of 4)+p(3hit out of 4)+p(All hit)
it is total probability-probability of not hitting yhr target
=>1-(3/4*3/4*3/4*3/4)
=>175/256

1 - P(he misses all four times)

Probability of a hit in one shot = 1/4
Probability of a miss in one shot = 1-1/4=3/4

1 - P(misses 1st time AND misses 2nd time AND misses 3rd time AND misses 4th time)

"AND" means "MULTIPLY":

1 - P(misses 1st time)*P(misses 2nd time)*P(misses 3rd time)*P(misses 4th time)


(1-3/4*3/4*3/4*3/4)=175/256

1........becoz every 4 shot 1 target is hit

how can he hit more than once when the question states he can hit only once in every 4 shots ?
please explain

What is the probability of hitting at least once? The event "at least one hit" could happen in several ways: (i) exactly 1 hit; (ii) exactly 2 hits; (iii) exactly 3 hits; (iv) exactly 4 hits.

(i) The probability of exactly one hit is (4C1)(1/4)(3/4)^3. This is because the hit could happen in any one of 4 (that is, (4C1) places).
Write H for hit and M for miss.
The probability of the pattern HMMM is (1/4)(3/4)(3/4)(3/4).
Similarly, the probability of MHMM is (3/4)(1/4)(3/4)(3/4).
You will notice this probability is the same as the probability of HMMM. We get the same probability for MMHM and for MMMH, for our total of (4C1)(1/4)(3/4)^3.

(ii) Similarly, the probability of exactly 2 hits is (4C2)(1/4)^2(3/4)^2.

(iii) The probability of 3 hits is (4C3)(1/4)^3(3/4).

(iv) The probability of 4 hits is (4C4)(1/4)^4.

Add up. We get the required answer.

However, that approach is a lot of work. It is much easier to find the probability of no hits, which is the probability of getting MMMM. This is (3/4)^4.
So the probability that the event "at least one hit" doesn't happen is (3/4)^4. So the probability that the event "at least one hit" does happen is 1−(3/4)^4.

Sorry for the long answer!!! :)

b)1/256 bcoz for 1 succession shot probability is 1/4
then for 4 succession shot probability must be 1/4*1/4*1/4*1/4 => 1/256

@AARUSHI MITTAL .because succession means 'sequence or series' , not 'hit successfully'.

d. 1- (3/4)^4

Ans-(175/256)

Option(D) is correct

The man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes.

So, the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes.

The probability that he will not hit the target in one shot =1 - Probability that he will hit target in exact one shot

=1−14=1−14

=34=34

Therefore, the probability that he will not hit the target in all the four shots

=(34)×(34)×(34)×(34)=(34)×(34)×(34)×(34)

=81256=81256
Hence, the probability that he will hit the target at least in one of the four shots:

=(1−81256)=(1−81256)

=175256