Great work, asax!asax wrote:UFF.. after 2 days of failing at BTG daily qs. i finally got one right!
Cheers,
Brent
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Great work, asax!asax wrote:UFF.. after 2 days of failing at BTG daily qs. i finally got one right!
It doesn't matter what order they pick the marbles in. The answer is the same.bluementor wrote:
I don't get it. Why do you assume that all the girls will pick the marbles first before the boys do or vice versa? If the girls and boys picked a marble each in turns, the result would be different.
Could anyone please explain? Thanks.
-BM-
Isn't there an issue with this question though. Our solutions generally assume that all of the girls necessarily choose first even though this is not explicitly stated in the problem. Wouldn't the answer change if we assumed that the girls and boys chose their marbles alternatively?Brent@GMATPrepNow wrote:The answer is, indeed, A (1/126)
Here's my solution:
The easiest/fastest way to determine the probability is to examine the probability of each necessary outcome to guarantee that the girls (and subsequently the boys) draw the same colored marble.
We get [P(1st girl selects any marble) x P(2nd girl selects marble the same color as 1st girl) x P(3nd girl selects marble the same color as 1st girl) x P(4th girl selects marble the same color as 1st girl) x P(5th girl selects marble the same color as 1st girl) x P(boys getting same color ball from remaining balls)
This equals: 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1
Which equals: 1/126
The order makes no difference. To demonstrate this, I'll refer you to jcnasia's post (6 posts before this one)bjc7227 wrote:Isn't there an issue with this question though. Our solutions generally assume that all of the girls necessarily choose first even though this is not explicitly stated in the problem. Wouldn't the answer change if we assumed that the girls and boys chose their marbles alternatively?Brent@GMATPrepNow wrote:The answer is, indeed, A (1/126)
Here's my solution:
The easiest/fastest way to determine the probability is to examine the probability of each necessary outcome to guarantee that the girls (and subsequently the boys) draw the same colored marble.
We get [P(1st girl selects any marble) x P(2nd girl selects marble the same color as 1st girl) x P(3nd girl selects marble the same color as 1st girl) x P(4th girl selects marble the same color as 1st girl) x P(5th girl selects marble the same color as 1st girl) x P(boys getting same color ball from remaining balls)
This equals: 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1
Which equals: 1/126
This solution would be correct if each selected marbles were replaced after each selection. That way, the probability would be 1/2 that a child chooses the correct color marble.smanstar wrote:What I did was
P( white marble) = 1/2
P ( black marble) = 1/2
So 5 girls selecting 5 white marbles and 5 boys selecting 5 black marbles
P = 1/2 * 1/2 * ...... 10 times = 1/2^10
Can someone explain where I am going wrong ???
Perfect!mparakala wrote:Ans: 1/126 (A)
This is how I solved it. Please let me know if I am correct!
5 girls can choose one set of same colore marbles in 5C5 ways or another set of 5 colored marbles in 5C5 ways
=> 5C5 +5C5 = 1+ 1= 2
now, the entire set of 10 marbles can be chosen by 5 girls in 10C5 = 10!/ 5!5! = 252
P(of choosing the same set of 5 marbles out of a total 10) = 2/252 = 1/126
Beautiful!sid128 wrote:I also got the same answer i.e. A (1/126) by applying following logic:
Total number of possibilities of distributing 5 white (identical) and 5 black (identical) marbles in 5 girls and 5 boys is (10!)/(5!)(5!)= 252
Favourable Events = 2 (Girls getting either all marbles or all black marbles)
Probability = No. of favourable events/No. of total events
= 2/252
= 1/126
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