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For one toss of a certain coin, the probability that the out

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Joined: 02 Oct 2009
Posts: 8
For one toss of a certain coin, the probability that the out  [#permalink]

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Updated on: 13 Jul 2013, 12:33
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Difficulty:

65% (hard)

Question Stats:

63% (02:07) correct 37% (02:08) wrong based on 269 sessions

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For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

Originally posted by ralucaroman on 25 Oct 2009, 17:38.
Last edited by Bunuel on 13 Jul 2013, 12:33, edited 2 times in total.
Renamed the topic, edited the question added the answer choices and OA.
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Joined: 02 Sep 2009
Posts: 50585
Re: Probability Question - GMATPrep  [#permalink]

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25 Oct 2009, 17:54
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3
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

$$P(h)=0.6$$, so $$P(t)=0.4$$. We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: $$P(h=5)=0.6^5$$;

4h and 1t: $$P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4$$, multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, $$P(h\geq{4})=0.6^5+5*0.6^4*0.4$$.

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Intern
Joined: 02 Oct 2009
Posts: 8
Re: Probability Question - GMATPrep  [#permalink]

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25 Oct 2009, 18:10
Hi, thank you for bringing light into this problem.

Your solution is correct, this is the gmatprep answer too: 5*(0.6)^4*(0.4) + (0.6)^5.

I had trouble understanding why the first part was multiplied with 5 (when there are 4 heads and 1 tail). I think I get it now (because 4H1T can happen in C5 taken by 4 times and 5H can happen in C5 taken by 5 which is 1). And addition of the 2 is due to fact that Probability(4H1T or 5H) = Probability (4H1T) + Probability (5H).
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Re: Probability Question - GMATPrep  [#permalink]

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25 Oct 2009, 18:12
1
Bunuel - the quant maestro --- nails one more problem... )

I missed the multiplying by 5 too
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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22 Mar 2015, 12:56
It's a straight forward E.
P (heads occurring 5 times) + P (heads occurring 4 times)P (tails occurring 1 time )

Only thing since the coin is tossed 5 times, it should be multiple by 5 - the second scenario.

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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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08 Dec 2017, 07:24
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: For one toss of a certain coin, the probability that the out &nbs [#permalink] 08 Dec 2017, 07:24
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