The probability of Jack hitting the bull’s eye is 40%. How many times

Question

The probability of Jack hitting the bull’s eye is 40%. How many times should he throw the dart in order to have more than 90% probability of hitting the bull’s eye at least once?

A. 3
B. 4
C. 5
D. 6
E. 7

Bunuel · Accepted Answer

For n throws the probability of hitting the bull’s eye at least once is 1 - P(not hitting at all in n throws) = 1 - (0.6)^n = 1 - (3/5)^n. We need this value to be greater than 90%:

1 - (3/5)^n > 9/10;

1/10 > (3/5)^n.

The least value of n is 5.

Answer: C.

Shobhit7 · Answer

Probability of hitting bull's eye per attempt= 0.4

Therefore, probability of not hitting bull's eye per attempt= 1-0.4= 0.6

The question is basically asking us to calculate number of attempts required to bring this "probability of not hitting bull's eye" to <=0.1

One Attempt, P(no bull's eye)=0.6

Two Attempts= 0.6*0.6=0.36

Three Attempts= 0.216

Four Attempts= 0.1296

So, we need five attempts to bring down "probability of not hitting bull's eye" to below 0.1

Hence, Ans C

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JeffTargetTestPrep · Answer

Since the probability of Jack’s hitting the bull’s-eye is 40%, the probability that he doesn’t hit the bull’s-eye is 60%, or 0.6. The probability that he will hit the bull’s-eye at least once is 1 - (probability that he will not hit the bull’s-eye)^n, where n is the number of times that he will not hit the bull’s-eye. Thus, we want:

1 - 0.6^n > 0.9

-0.6^n > -0.1

0.6^n < 0.1

Since 0.6^4 = 0.1296 and 0.6^5 = 0.07776, n = 5.

Answer: C

delid · Answer

Hello  Bunuel , thanks for the explanation. I understand the logic, but once you have identified that the key is to find n so that 1/10 > (3/5)^n ... is there a quick shortcut for doing the calculations that i am not seeing?

Thanks a lot for your support!

Bunuel · Answer

You can find it by trial and error.

1/10 is an easy fraction for comparison and 3/5 = 6/10, so it should not be hard to do.

dyogendrarao · Answer

Hi Bunuel,

What if I solve like this:

BE = Bull's eye
N = number of attempts
P(BE) x N x P(not BE) > 0.9
0.4 x N x 0.6 > 0.9
N > 4.7

Therefore, the answer is N=5.

Do you see any flaw in this method?

CrackverbalGMAT · Answer

The probability of Jack hitting the bull’s eye is 40%. How many times should he throw the dart in order to have more than 90% probability of hitting the bull’s eye at least once?

Probability of Jack hitting the target is 40 % i.e 40/100 = 2/5

Probability of Jack not hitting the target = 1- 2/5 = 3/5

In 'n' throws, P( jack hitting the bulls eye at least once) = 1- P( Jack not hitting the bull eyes in n throws)= 1 - (3/5)^n

We need to find the value of n , such that P( jack hitting the bulls eye at least once) is greater than 90%

1 - (3/5)^n > 90/100

1-9/10 > (3/5)^n

(3/5)^n < 1/10
(.6)^n <.1

n= 5 is the min value that satisfies the condition.

Option C is the answer.

Thanks,
Clifin J Francis,
GMAT SME

bumpbot · Answer

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The probability of Jack hitting the bull’s eye is 40%. How many times

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