ggarr wrote:
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?
A. 2/5
B. 3/5
C. 1/18
D. 1/3
E. 1/6
This is a conditional probability question. The formula for conditional probability says something like -
P (B/A) = P (A and B) / P(A) [Read as Prob. of event B given event A has already occured] A: event that sum of the numbers on the 2 cards is 8
B : event that one of the 2 cards is numbered 5
Whats P(A)?favorable outcomes = (2,6) (3,5) (5,3) (6,2) (4,4)
n(favorable outcomes) = 5
n(total outcomes) = 6C2 = 15
P(A) = 5/15 = 1/3
Whats P(A and B) ?It means sum is 8 and one of the cards is numbered 5
Well, that's only 2 cases - (5,3) (3,5)
n(favorable outcomes) = 2
n(total outcomes) = 6C2 = 15
P(A and B) = 2/15
Substitute in formula P(B/A) = Prob. that one of the 2 cards is 5 given their sum is 8
= (2/15)/(1/3)
= 2/5 Ans.