- #1
erisedk
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Homework Statement
Consider a bag containing 10 balls of which a few are black balls. Probability that bag contains exactly 3 black balls is 0.6 and probability of bag containing exactly 1 black ball is 0.4. Now, balls are drawn from the bag, one at a time, without replacement, till all black balls have been drawn. The probability that this procedure would end at the 6th draw is p
Homework Equations
The Attempt at a Solution
If bag contains 3 black balls, P(6) = 0.6 * (3.2.7.6.5.1)/(10.9.8.7.6.5) * 5! = 0.6 (5! because that's the possible ways of arranging 3, 2, 7, 6, 5 in different orders)
If bag contains 1 black ball, P(6) = 0.4 * (9.8.7.6.5.1)/(10.9.8.7.6.5) = 1/25
So, P(6) = 0.6 + 1/25 = 16/25
Which is wrong.