There are 3 cards in a box. One card is blue on both sides, another is

Question

There are 3 cards in a box. One card is blue on both sides, another is red on both sides, and the last is red on one side and blue on the other. One cards is picked at random from the box. If the face up side shows blue, what is the probability the other side, the face down side, is also blue?

A. 5/6
B. 3/4
C. 2/3
D. 1/2
E. 1/3

A. 5/6
B. 3/4
C. 2/3
D. 1/2
E. 1/3

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chetan2u · Accepted Answer

We can use conditional formula too, but we can easily do this by some logical visualization

We are given that the face up is BLUE.
So the card has to be either BB or BR. Now it is easy to take 1/2 as the chance of each card.

But we are looking at Blue colour. Let the sides be 1 and 2, so  in BB, there are 2 ways the side Blue can be chosen - B_1  or  B_2 , and  in BR, there is one way that B can be chosen.

Thus, there are 3 different possibilities of B, that is face up when picked up - it could be  B_1, B_2  or B, and in two cases -  B_1, B_2  -the opposite side is B.
Probability =  \frac{2}{3} .

C

Babineaux · Answer

Let the cards be like,

BB..............RR..............BR

Now one of the card has already chosen and up-face is blue,

Then whole thing comes to two choices, _B.............._R.

"_" represent already chosen.

Thus the probability that the down-face is blue will be 1/2.

IMO ans is D.

dimri10 · Answer

i will go with C the nagative approach. if you picked a blue one ( it means it is either the blue/blue card or the blue/red cars), so the chances that the other side is blue is equal to 1- blue/red card.
picking a blueqred card out of 3 cards equals to 1/3 so, i hope, it should be 1-1/3. or 2/3

Lipun · Answer

Hi  chetan2u ,

Please confirm if my understanding and approach are correct.

Conditional Probability: P(A & B) = P(A) * P(B|A).
We need to find out P(B|A).

Once we pick a BB card, both sides are always going to be blue. So, probability of face-up blue = 1/3 * 1 = 1/3
Once we pick a BR card, the probability of face-up blue = 1/3 * 1/2 = 1/6
P(A) is the event that the face-up color of the card is blue = P(BB) + P(B in BR) = 1/3 + 1/6 = 1/2

P(A & B) is the event that we picked a BB card = 1/3

So, P(B|A) = (1/3)/(1/2) = 2/3

Thanks

PS: I somehow find it similar to the Monty Hall problem but cannot relate it exactly how.

chetan2u · Answer

Yes, you are correct with the conditional probability method.

GMATinsight · Answer

Total faces on 3 cards = 6
Total Blue faces on 3 cards = 3

Probability of selecting a card and seeing it's face blues = 3/6 = 1/2

Total Cards = 3
Favourable card with both faces blue = 1

Probability of picking a card which has both sides blue = 1/3

Required probability =  (1/3) / (1/2)  = 2/3

Answer: Option C

sambitspm · Answer

This is straightforward.
One side is blue. Now we need to find the probability that other side is also blue.

P(blue) = 2 blue / 3 cards
2/3
IMO C

Shobhit7 · Answer

Bunuel 
 chetan2u

Please advise whether my below stated understanding is correct.

Event 1: Blue face
Total 6 faces, 3 red and 3 blue
If event 1 is unknown: P(E1) = 3/6 = 1/2
If event 1 is known and favorable: P(E1) = 1

Event 2 given Event 1: Blue face
As 1st event is known and is a blue face, now
Total possible unknown faces are 3: 2 Blue and 1 Red
Therefore, P(E2) = 2/3

Therefore P(E2 given E1)= 1*2/3 = 2/3

Hence, Ans C

Srinivas0512 · Answer

Let there be two cases.

Case 1 : B/B B/R R/R
In this case we have two blue cards face up.
 
The probability of selecting a blue card is thus 2/3 and the probality of the face down color being blue is 1/2. Thus the probability that we have a card blue on both sides is 2/3*1/2 = 1/3.

Case 2 : B/B R/B R/R
In this case we have just one card which face up color blue. And thus the probabiliy of selecting this card is 1/3. And since this is the only card that has face down color blue, the probability of selection is 1. Thus the probability that we have a card blue on both sides is 1/3*1 = 1/3.

Total Probability is P(Case1) + P(Case2) = 1/3+1/3 = 2/3.

IanStewart · Answer

There are 3 equally likely ways to pick a blue side. In 2 of those ways, the opposite side is also blue, so the answer is 2/3.

IanStewart · Answer

The question in the OP is actually a famous problem, similar to the Monty Hall problem, known as "Bertrand's Box Paradox". In Bertrand's version, there are three boxes, one with two gold coins, one with two silver coins, and one with one coin of each type, but conceptually it's the same question as the cards question in this thread.

mSKR · Answer

i am not able to visualize in practical manner.

What if the question is changed to: what is the probability the other side, the face down side, is red?
would it be 1/3 or 2/3? why or why not?

ScottTargetTestPrep · Answer

Solution:

We can label each of the sides of the cards as follows:

Card 1: B1, B2

Card 2: R1, R2

Card 3: B3, R3

Since we know the side that shows facing up is blue, it could be B1, B2, or B3. If it’s B1, the facing down side is B2. If it’s B2, the facing down side is B1. If it’s B3, the facing down side is R3. We see that of the 3 possible facing down sides, 2 of them are blue. Therefore, the probability is 2/3.

Answer: C

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There are 3 cards in a box. One card is blue on both sides, another is

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