Is it 1/3?
Three cards are in a hat. One is red on both sides, one is white on both sides, and one is red on one side and white on the other. I draw a card from the hat, and drop it on the table. The upward-facing side is red. What are the odds that the downward-facing side is also red?
- None of the other options is correct.
- Fifty percent.
- Two out of three.
- One out of three.
Ok. I have always had problems with this type of question. I would go with CTG for conditional probability, but I have seen where the sample space reduces because of a given value.
As yoda says,
Probabilty that given card is Red = 1 (as it is already given)
Does it mean that the sample space is reduced to 2 hats (with red) as against 3 hats?
Or should it be
P(Red hat) x P(both red given that the first one is red) = 2/3x1/2 = 1/3
Thanks
My answer was 50% since the sample space would reduce to 2 hats.
I do not have the right answer for this.
However, let me check this with Dr.Math and I will get back to you guys.
This is the reply that I got from Math forum:
Your confusion stems from the fact that there are in fact two random
events that occur in the display of a color: (1) The selection of one
of three cards, and (2) the selection of which side of the card is shown.
To compute the correct probability, let us label the cards as follows:
{r1, r2}
{r3, w1}
{w2, w3}.
Clearly, each of the faces r1, r2, r3, w1, w2, w3 have equal
probabilities of 1/6 of being shown. Then in how many of the 3 out
of 6 ways to show a red card face-up is the face-down side also red?
We enumerate the possibilities as follows, where the first value in
a given pair denotes the face-up color, and the second value denotes
the face-down color:
{r1, r2}
{r2, r1}
{r3, w1}.
Therefore, the probability that the face-down color is red given that
the face-up color is red is simply 2/3, which is neither 1/2 nor 1/3.
Last edited by pach2212; 01-03-2006 at 08:41 AM. Reason: Automerged post
CTG,
In the reasoning provided by Dr. Math, he gives different names for different sides of the coins. Since we have 3 coins, there are 6 diffferent names.
Red-Red coin = r1 and r2
Red-White coin = r3 and w1
White-White coin = w2 and w3
It is given that, when a coin is chosen and placed on the table, the face pointing upwards is Red.
So this face can be "r1 or r2 or r3." ------------- (1)
Accordingly we will have the face pointing downwards to be
"r2 or r1 or w1" ------------------------ (2) (The other faces of the coins listed in (1) )
Now, the question asks the odds that the downward face is also Red.
So from (2) we have 2 red faces out of 3.
Hence the answer, 2/3.
Let me know if it is unclear
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