A bag contains red and blue marbles totalling between 50 and 100 in number. If two marbles are drawn at random, the chances of them both being red are 25%. One third of the marbles are picked from the bag, at random, and thrown away. If one marble is now drawn at random from the bag, what is the probability of being blue ?
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75\100 is the probability of being blue in my opinion.
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Probability is 1/2
Let us say initially there are B number of Blue marbles and R number of Red marbles.
given 25% = 1/4 = R/(B+R) * (R-1)/(B+R-1)
As B and R are large, B = R and the probability is not exactly equal to 25%. It is only approximate value.
Total number of marbles = 2N. Blue = N, Red = N.
When 2N/3 marbles are randomly taken out, 4N/3 marbles are left in the bag. Of them the number of the blue marbles could be:
N/3, N/3+ 1, N/3 + 2, N/3 +3, ....., N-1 or N
There are 1 + 2N/3 number of outcomes. Probability of each = 1/[1+ 2N/3]
Probability of the one ball drawn now being blue:
1/[1+ 2N/3] *[ (N/3)/(4N/3) + (1+N/3)/(4N/3) +......+ N/(4N/3) ]
= 1/[1+ 2N/3] * [1 + 2N/3] * [N/3 + N] / 2 * 1/(4N/3)
= 1/2
Let us say initially there are B number of Blue marbles and R number of Red marbles.
given 25% = 1/4 = R/(B+R) * (R-1)/(B+R-1)
As B and R are large, B = R and the probability is not exactly equal to 25%. It is only approximate value.
Total number of marbles = 2N. Blue = N, Red = N.
When 2N/3 marbles are randomly taken out, 4N/3 marbles are left in the bag. Of them the number of the blue marbles could be:
N/3, N/3+ 1, N/3 + 2, N/3 +3, ....., N-1 or N
There are 1 + 2N/3 number of outcomes. Probability of each = 1/[1+ 2N/3]
Probability of the one ball drawn now being blue:
1/[1+ 2N/3] *[ (N/3)/(4N/3) + (1+N/3)/(4N/3) +......+ N/(4N/3) ]
= 1/[1+ 2N/3] * [1 + 2N/3] * [N/3 + N] / 2 * 1/(4N/3)
= 1/2
kvnmurty:
click on red heart thanks above pls
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