Set-7; Q-14

[kannn]

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

[PennMI]

is it D?

[eltonr]

It should be C)
Prob (One of cards is 5) = P(1st no = 5) + P(2nd no = 5) = 1/6 + 1/6 = 1/3

[dwurster]

I would also say that it should be D.), but I'm not sure...
I just want to write down my thoughts. Please correct me if I do a stupid mistake:

We know that the sum of the two card numbers is 8. According to this, we have the following options which fulfill this condition:

first card, second card
2,6
6,2
3,5
5,3
4,4
(these are the possible events)

the probability that the first card is a 5 is 1/5
the probability that the second card is a 5 is also 1/5
so the probability that one of the two cards is a 5 is 2/5?!?

(the card which is a 1 could have never been drawn, because there is no way to get 8 with it...?)


did I fall in a trap?

[sjd00d]

D. 2/5 for me too. What's the OA

Dwurster,

I did it using similar approach too!!

[kannn]

Originally Posted by dwurster

I would also say that it should be D.), but I'm not sure...
I just want to write down my thoughts. Please correct me if I do a stupid mistake:

We know that the sum of the two card numbers is 8. According to this, we have the following options which fulfill this condition:

first card, second card
2,6
6,2
3,5
5,3
4,4
(these are the possible events)

the probability that the first card is a 5 is 1/5
the probability that the second card is a 5 is also 1/5
so the probability that one of the two cards is a 5 is 2/5?!?

(the card which is a 1 could have never been drawn, because there is no way to get 8 with it...?)


did I fall in a trap?

OA is D. Thanks. I counted the card which has 1 and chose the wrong answer i.e 1/6 + 1/6 = 2/6 = 1/3 which is wrong.

[mathphobia]

dwurster

the probability that the first card is a 5 is 1/5
the probability that the second card is a 5 is also 1/5
so the probability that one of the two cards is a 5 is 2/5?!?

(the card which is a 1 could have never been drawn, because there is no way to get 8 with it...?)


did I fall in a trap?[/quote]

dwurster why do you say the probability of 1st card is 1/5 and not 1/6

[dwurster]

mathphobia:

we don't have to consider probabilities of getting a sum of 8
-> it is fact (!) that we have two cards which give us a sum of 8!
-> so there are only 5 cards which come into question (2,3,4,5,6) because the card 1 has no complement which could yield 8! so we just have 5 different possible cards to consider.

all possibilities which fulfil the sum of 8 are:
2,6
6,2
3,5
5,3
4,4

now you have two ways to get the answer:
1.) add the two probabilities of getting a 5 when drawing two times (considering that all cards we could ever have are 2,3,4,5 and 6 (not 1))
-> 1/5 + 1/5

2.) divide the number of events where a 5 is given [ (5,3) + (3,5) = 2] by the number of all possible events [ ( 2,6) + (6,2) + (3,5) + (5,3) + (4,4) = 5]
-> 2/5