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

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

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22 Oct 2017, 22:36
1
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Difficulty:

65% (hard)

Question Stats:

46% (01:47) correct 54% (01:53) wrong based on 105 sessions

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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

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Joined: 02 Sep 2009
Posts: 50670
Re: The probability of Jack hitting the bull’s eye is 40%. How many times  [#permalink]

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23 Oct 2017, 00:02
3
nkmungila wrote:
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

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.

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Re: The probability of Jack hitting the bull’s eye is 40%. How many times  [#permalink]

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26 Oct 2017, 14:31
nkmungila wrote:
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

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.

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Re: The probability of Jack hitting the bull’s eye is 40%. How many times  [#permalink]

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30 Oct 2017, 13:07
Bunuel wrote:
nkmungila wrote:
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

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.

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!
Math Expert
Joined: 02 Sep 2009
Posts: 50670
Re: The probability of Jack hitting the bull’s eye is 40%. How many times  [#permalink]

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30 Oct 2017, 22:39
1
1
delid wrote:
Bunuel wrote:
nkmungila wrote:
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

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.

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!

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.
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Joined: 01 Feb 2017
Posts: 167
Re: The probability of Jack hitting the bull’s eye is 40%. How many times  [#permalink]

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30 Nov 2017, 13:17
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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Re: The probability of Jack hitting the bull’s eye is 40%. How many times &nbs [#permalink] 30 Nov 2017, 13:17
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