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A coin is weighted so that the probability of heads on any

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by AAPL » Wed Nov 06, 2019 2:14 am
Manhattan Prep

A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. \(0.6^5+5(0.6^4)(0.4)+10(0.6^3)(0.4^2)\)
B. \(0.6^5+4(0.6^4)(0.4)+6(0.6^3)(0.4^2)\)
C. \(0.6^5+3(0.6^4)(0.4)+2(0.6^3)(0.4^2)\)
D. \(0.6^5+2(0.6^4)(0.4)+(0.6^3)(0.4^2)\)
E. \(0.6^5+(0.6^4)(0.4)+10(0.6^3)(0.4^2)\)

OA A
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weighted coin

by GMATGuruNY » Wed Nov 06, 2019 3:13 am
AAPL wrote:Manhattan Prep

A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. \(0.6^5+5(0.6^4)(0.4)+10(0.6^3)(0.4^2)\)
B. \(0.6^5+4(0.6^4)(0.4)+6(0.6^3)(0.4^2)\)
C. \(0.6^5+3(0.6^4)(0.4)+2(0.6^3)(0.4^2)\)
D. \(0.6^5+2(0.6^4)(0.4)+(0.6^3)(0.4^2)\)
E. \(0.6^5+(0.6^4)(0.4)+10(0.6^3)(0.4^2)\)
ALWAYS LOOK AT THE ANSWER CHOICES.

P(no more than two tails) = P(all heads) + P(4 heads and 1 tails) + P(3 heads and 2 tails)

P(heads) = 0.6
P(tails) = 0.4
Thus, the second term in each answer choice must represent P(4 heads and 1 tails):
(0.6)�(0.4)

Since the 1 tails can happen on any of the 5 flips -- implying 5 different ways to get exactly 1 tails -- the second term must include a factor of 5:
5(0.6)�(0.4)
Only A satisfies this condition.

The correct answer is A.
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by Scott@TargetTestPrep » Thu Nov 07, 2019 6:37 pm
AAPL wrote:Manhattan Prep

A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?

A. \(0.6^5+5(0.6^4)(0.4)+10(0.6^3)(0.4^2)\)
B. \(0.6^5+4(0.6^4)(0.4)+6(0.6^3)(0.4^2)\)
C. \(0.6^5+3(0.6^4)(0.4)+2(0.6^3)(0.4^2)\)
D. \(0.6^5+2(0.6^4)(0.4)+(0.6^3)(0.4^2)\)
E. \(0.6^5+(0.6^4)(0.4)+10(0.6^3)(0.4^2)\)

OA A
No more than two tails means:

1) 0 tail and 5 heads, 2) 1 tail and 4 heads, and 3) 2 tails and 3 heads

Now let's determine the probability of each case above:

1) P(0 tail and 5 heads) = P(HHHHH) = 0.6^5

2) P(1 tail and 4 heads) = P(HHHHT) x 5!/4! = 0.6^4 x 0.4 x 5

3) P(2 tails and 3 heads) = P(HHHTT) x 5!/(3! x 2!) = 0.6^4 x 0.4^2 x 10

Thus, the overall probability is:

0.6^5 + 5 x 0.6^4 x 0.4 + 10 x 0.6^3 x 0.4^2

Answer: A

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