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The probability of a shooter hitting a target is 34\frac34. How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 90%90 \% ?

Solution

✅ Correct Option: 2

For n=1n = 1 shot:

Probability of at least one hit =1−= 1 - Probability of missing

=1−(0.25)1= 1 - (0.25)¹

=0.75= 0.75

For n=2n = 2 shots:

Probability of at least one hit =1−= 1 - Probability of missing all shots

=1−(0.25)2= 1 - (0.25)²

=1−0.0625= 1 - 0.0625

=0.9375= 0.9375

Since 93.75%>90%93.75\% > 90\%, the minimum number of shots needed is 2.

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