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

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22 May 2006, 14:13
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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(0.4)
D) 4(0.6)^4(0.4) + (0.6)^5
E) 5(0.6)^4(0.4) + (0.6)^5
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22 May 2006, 14:27
Shouldn't the probability be

0.4*(0.6^4) + 0.6^5
=0.6^4 (1)
=0.6^4

Hopefully that is (B) amognst the options. Also looks like the correct options are
A) (0.6)^5
B) (0.6)^4
C) (0.6)^4(0.4)
D) (0.6)^4(0.4) + (0.6)^5
E) (0.6)^4(0.4) + (0.6)^5
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22 May 2006, 14:44
P(HHHHT) + P(HHHHH) =

(0.6^4 * 0.4 ) + (0.6^5)
= 0.6^4 (0.4 +0.6) =

= 0.6^4
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22 May 2006, 15:51
The OA is GMATPrep is E.
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22 May 2006, 16:24
remgeo wrote:
P(HHHHT) + P(HHHHH) =

(0.6^4 * 0.4 ) + (0.6^5)
= 0.6^4 (0.4 +0.6) =

= 0.6^4

I got a similar approach as remgeo, but
P(HHHHT) = will have 5C4 and P(HHHHH) = there is only one way to get this.
P(HHHHT)+ P(HHHHH) =5C4 * (0.6^4 * 0.4 ) + 0.6^5
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22 May 2006, 17:38
john2005 wrote:
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(0.4)
D) 4(0.6)^4(0.4) + (0.6)^5
E) 5(0.6)^4(0.4) + (0.6)^5

5C4(0.6)^4(0.4)^1 + 5C5(0.6)^5(0.4)^0
=5(0.6)^4+(0.6)^5
Thus E
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22 May 2006, 18:42

HHHHT --> 5!/4!1! = 5 ways
HHHHH --> 1 way

For HHHHT, each permutation has a probability (0.6)^4(0.4), 30 such permutations will have a probability 5(0.6)^4(0.4)

For HHHHH, the probability = (0.6)^5

Total = 5(0.6)^4(0.4) + (0.6)^5
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23 May 2006, 07:06
Great question. I initially forgot to account for for the different ways to get HHHHT (5C4).
23 May 2006, 07:06
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