A box contains four coins, of which two . . .

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A box contains four coins, of which two . . .

by Vincen » Thu Sep 14, 2017 2:26 pm
A box contains four coins, of which two coins have heads on both their faces, one coin has tail on both its faces and the fourth coin is a normal one. A coin is picked at random and then tossed. If head is the outcome of the toss, then find the probability that the other face (hidden face) of the coin tossed is also a head.

A. 2/5
B. 1/2
C. 4/5
D. 2/3
E. 3/4

The OA is C.

This is a difficult question. I need help, please.

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by GMATGuruNY » Thu Sep 14, 2017 6:57 pm
Vincen wrote:A box contains four coins, of which two coins have heads on both their faces, one coin has tail on both its faces and the fourth coin is a normal one. A coin is picked at random and then tossed. If head is the outcome of the toss, then find the probability that the other face (hidden face) of the coin tossed is also a head.

A. 2/5
B. 1/2
C. 4/5
D. 2/3
E. 3/4
Two coins have heads on both faces:
H, H
H, H
One coin has heads on only one face:
H, T.

One coin has tails on both faces:
T, T.

Since the outcome of the toss is heads, the flipped coin must be one of the 3 coins in blue.
The 3 coins in blue comprise a total of 6 faces:
H, H
H, H
H, T.

Since the outcome of the toss is heads, one of the H's above is facing upward on the toss.
Thus, one of the 5 remaining faces -- H, H, H, H, T -- must be facing downward.
Of these 5 remaining faces, 4 are heads.
Thus, P(downward face is heads) = 4/5.

The correct answer is C.
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by [email protected] » Fri Sep 15, 2017 9:20 am
Hi Vincen,

This question can be approached with a bit of 'brute force' math - you just have to write everything down and stay organized. To start, I'm going to refer to 8 faces on the 4 coins as:

H1/H2
H3/H4
H5/T1
T2/T3

Based on the information in the prompt, we know that 5 of the 8 'faces' are heads, so if we randomly pick a coin and flip 'heads', then one of the first 3 coins MUST have been chosen (since the 4th coin has two 'tails', it could not ever flip 'heads').

From there, we're asked for the probability that the other side of the coin is also 'heads.' With a 'heads' face-up, here are the following possible outcomes:
H1 up/H2 down
H2 up/H1 down
H3 up/H4 down
H4 up/H3 down
H5 up/T1 down

4 of the 5 'heads up' options have a 'head' on the other side.

Final Answer: C

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Vincen wrote:
Thu Sep 14, 2017 2:26 pm
A box contains four coins, of which two coins have heads on both their faces, one coin has tail on both its faces and the fourth coin is a normal one. A coin is picked at random and then tossed. If head is the outcome of the toss, then find the probability that the other face (hidden face) of the coin tossed is also a head.

A. 2/5
B. 1/2
C. 4/5
D. 2/3
E. 3/4

The OA is C.

This is a difficult question. I need help, please.
Since there are only 3 coins that have a head on at least one face and the outcome of the toss is a head, we can ignore the coin with tails on both faces. Now, let’s label the faces of the other 3 coins.

One of coins with both heads: H1, H2

The other coin with both heads: H3, H4

The coin with one head and one tail: H5, T1

Since the outcome of the toss is a head, it’s possible is from one of the following (shown face / hidden face):

H1 / H2

H2 / H1

H3 / H4

H4 / H3

H5 / T1

We see that from the 5 outcomes above where the shown face is a head, 4 of them have a head on the hidden face also. Therefore, the probability that the hidden face of the coin tossed is also a head is 4/5.

Answer: C

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