Problem Solving

Q10:
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 ?
F. 1/6
G. 1/5
H. 1/3
I. 2/5
J: 2/3

FGH....mean ABCDE...it was like that only?

this is sort of a conditional probability

total permutations of = 6P2 =15

now out of this 5 i.e 62&vice versa, 53&vv &44 will satisfy the event that sum is 8 & out of this in two cases on of the card is 5

so reqd P =2/5

Hi Samir,
The second part of the answer is pretty logical.Can you pls explain why 6P2 and how??

Hey Maxx - according to me, that 6P2 what Samir has mentioned in his post, is not at all required. We have to take care of the total possibilities and also the favorable occurances that you can get what Samir has mentioned later.

Hi Amitava,
Well its certainly not required on ur scratch paper, but I think it is important to convey that the stmt "sum of the numbers on the cards is 8" shrinks the original sample space(total outcomes) from 15 to 5, so I thought it would be good to proceed in a logical order here(for the expln), though it is not mandatory.

why not 2/15....? 2 is the favorable outcome meeting all the cond....

I understand what you are trying to say Amitava.Thanks!!

Well neeraj 2/15 is not possible because as Samir had mentioned, the total number of possibilites is 5(62&vice versa, 53&vv &44) and the probability of getting a 5 on the other card is 2 and so 2/5.

Am I right guys??

> total permutations of = 6P2 =15
Samir,
Not that this number is used in the problem, but 6P2 is 30, not 15.

Thanks Naren, I think it was a silly mistake, anyways it never mattered for this problem, just use to explain the concept. Thanks for pointing out though.

[quote="samirpandeyit62"]Thanks Naren, I think it was a silly mistake, anyways it never mattered for this problem, just use to explain the concept. Thanks for pointing out though.[/quote]

Samir

The cards are drawn after replacement. For the sake of generalization - shdnt it be 36 total outcomes

Hi rprasanna,
The question uses conditional probabilty so ya we may consider the total nos of outcomes but it can be anything say "n", it wont really matter for calculating the probabilty though.