One coin in a collection of 65 has two heads. The rest are fair. If a

Question

One coin in a collection of 65 has two heads. The rest are fair. If a coin, chosen at random from the lot and then tossed, turns up heads 6 times in a row, what is the probability that it is the two-headed coin?

A. 1/65
B. 1/64
C. 1/3
D. 1/4
E. 1/2

Are You Up For the Challenge: 700 Level Questions

A. 1/65
B. 1/64
C. 1/3
D. 1/4
E. 1/2

Are You Up For the Challenge: 700 Level Questions

imeanup · Accepted Answer

Probability of selecting 2-headed coin,  P(A) =\frac{1}{65}

Probability of choosing fair coin:  P(X) = 1 – P(A) = 1 - \frac{1}{65}=\frac{64}{65}

For a 2 headed coin, the probability of flipping 6 heads in a row:  \frac{6}{6}

For a fair coin, the probability of flipping 6 heads in a row:  \frac{1}{2^6}  or  \frac{1}{64}

If we don’t know which coin was chosen, then the probability of flipping 6 heads in a row  P(B) ;

P(B) = \frac{1}{65}*\frac{6}{6} + \frac{64}{65}*\frac{1}{64}=\frac{1}{65}+\frac{1}{65}=\frac{2}{65}

Combining the probabilities of P(A), which is selecting the 2 headed coins, and P(B), which is all 6 tosses land as heads.

P(\frac{A }{B}) = \frac{P(A) }{ P(B)} * \frac{6}{6 }= \frac{1*65}{2*65} =\frac{1}{2}

IMO  Ans E

Ekland · Answer

The probability of choosing the two headed coin from the lot is 1/65. This is because there are 65 coins and the two headed coin is just one out of the 65. It doesn't matter what you did with the coin after choosing it. That's not part of the math in any sort of way. It's just there to distract u.
A is the answer

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CrackverbalGMAT · Answer

SOLUTION: The Probability of selecting a fair coin = 64/65 The Probability of getting 6 heads in a row with the fair coin = (1/2)^6= 1/64 The Probability of selecting the unfair 2-headed coin = 1/65 Probability of getting 6 heads in a row with the 2-headed coin = 1 If a coin, chosen at random from the bag and then tossed, turns up heads 6 times in a row, the probability that it is the two-headed coin(P (B/A)) = / = (1/65)/(2/65) = 1/2 Hence OPTION (E) EXPLANATION: conditional probability, Bayes' theorem of P(B|A)= P(B) P(A|B) / P(A) Where A has already occurred(we're getting 6 heads in a row) & B is the event we are interested in to calculate(coin is a 2 headed coin) when A has already occurred. P(A/B)=Probability of getting 6 heads in a row, if coin is 2-headed= 1 P(B)=Probability that the coin is two- headed =1/65 P(B/A)=Probability of B, if A has already occurred P(A)=Probability of getting 6 heads in a row=64/65)*(1/64)}+{(1/65)*1} ->Fair coin selected * probability to get 6 heads from them + two headed selected * probability to get 6 heads from it-] Hope this helps

Devmitra Sen(Math)

spiedrar101 · Answer

This is a conditional probability question.

We know that P(A|B) = P(A and B)/P(B). We want the probability that A = the coin is loaded, given that B = we got 6 heads in a row.

First P(B) = P( 6 heads in a row) = (1/2^6)*(64/65) + 1*(1/65) = 2/65. This is a weighted probability.

Now, the probability that A is loaded and that we got six heads is just the probability that A is loaded, since if it is loaded we must have all heads. Thus, P(A and B) = P(A) = 1 / 65.

Putting it all together, P(A|B) = P(A and B)/P(B) = (1/65) / (2 / 65) = 1/2

Hence, E is the correct answer

Gknight5603 · Answer

baye's theorem...

1/(64C1*2^-6+1C1*1)
= 1/2

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bumpbot · Answer

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One coin in a collection of 65 has two heads. The rest are fair. If a

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