adityaganjoo wrote:
Bunuel wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8?(A) 1
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{2}{5}\)
(E) \(\frac{1}{3}\)
What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened.
Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.
Answer: D.
Check:
https://gmatclub.com/forum/the-probabili ... 85523.html for more.
Hope it's clear.
Bunuel Requesting for your help here! This is my weakest area
Please tell me why am I wrong?
Facts:
(i) We select numbers at random
(ii) Only possible ways to get a sum of 8 in this situation are: (5,3) & (3,5)
Now. there are 2 cases possible:
Case 1: 5 is selected first, then 3
Probability = (1/6)*(1/6) = 1/36
Case 2: 3 is selected first, then 5
Probability = (1/6)*(1/6) = 1/36
Total Probability = (1/36) + (1/36) = 1/18
Why is this wrong?
QUESTION 1: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that the sum of the two cards will be 8?P(sum of 8) = P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) + P(4, 4) = 1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 = 5/36.
QUESTION 2: There are 6 cards numbered from 1 to 6. They are placed into a box, and then two cards are drawn without replacement. What is the probability that the sum of the two cards will be 8?P(sum of 8) = P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) = 1/6*1/5 + 1/6*1/5 + 1/6*1/5 + 1/6*1/5 = 2/15.
Here (4, 4) case is not possible since we have only one 4 in the box and we are not putting back cards we draw.ORIGINAL QUESTION: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 IF the sum of the two was 8?This one is different from the two above because We are not asked to find the probability of getting the sum of 8, we are told that the sum of 8 has already occurred -
we have the sum of 8. So, the question asks: if we have the sum of 8, what is the probability that one of the cards was 5?
Now, considering all above check the solution provided.
What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened.
Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.
Answer: D.
Does this make sense?