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. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
I don’t think that the answer is listed, correst me if I’m wrong:
probability of drawing 5 in the first round is 1/6, probability of drawing 3 in the second round is also 1/6 so the total probability of getting one 5 is 1/6 times 1/6 = 1/36, plus counting that you can draw 3 and then 5 it is 1/36 + 1/36 = 1/18
the answer is 1/18 i think
D
for sum to be 8
we can have (5,2) (2,5) (1,6) (6,1) (4,4)
and for having 5 we have 2 cases which is (2,5) (5,2)
hence 2/5
“C”
The card can be selected 1st or second, hence you must come up with all possibilities. (4,4),(3,5),(5,3),(2,6),(6,2)…
Two of the 5 options contain a 5 so answer is 2/5…
Ans is D
For sum to be 8, we can have only 5 possibilities : (3,5),(5,3) (2,6), (6,2) and (4,4)
Prob. of drawing 2 cards one of which is 5 is 2/5
didn’t read carefully:P
answer is D indeed
I think it is “C’ -1/3.
Different possibilities of sum tobe 8 are,
2,6 / 4,4 / 3,5 / 5,3 / 6,2.
possibility of 5 is 2 times out of 5 .
so answer is 2/5
“D”
D
the answer appears to be D:
D
I will go with option D)2/5
since cards are drawn randomly u can’t make sure that sum wd be 8, therefore total possible events sd 36 instead of 5. I think 1/18 is the correct answer.
It´s …. 1 / 18…. but if since options are given so i go for 2 / 5.
D…
Principle of composed probabilities
Probability(one card =5 if ?two cards=8) = Probability(?two cards=8) x Probability(one card =5/?two cards=8)
Probability(?two cards=8) = Number of wanted outcomes / Total Number of outcomes
Number of wanted outcomes = 5 which are (2;6) (3;5) (4;4) (6;2) (5;3)
Total Number of outcomes = 36 equal to 6 outcomes times 6 outcomes
Probability(?two cards=8) = 5/36
Probability(one card =5/?two cards=8) = Number of wanted outcomes / Total Number of outcomes
Number of wanted outcomes = 2 which are (3;5) and (5;3)
Total Number of outcomes = 5
Probability(one card =5/?two cards=8) = 2/5
So Probability(one card =5 if ?two cards=8) = (5/36)*(2/5) = 2/36 = 1/18
In fact Probability(one card =5 if ?two cards=8) = Probability(5 and 3)
SINCE 1/18 IS NOT LISTED WE SHOULD UNDERSTAND THAT the DISERED PROBABILITY IS Probability(one card =5/?two cards=8) = 2/5 AND THEREFORE correct answer SHOULD BE D.
I think its C,
No of possible outcomes : 53,35,62,26,44,44
No of favored outcomes : 53,35
Probability : 2/6 -> 1/3