An urn contains 10 black and 5 white balls. T | vedclass

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(A) Let $E$ and $F$ denote respectively the events that the first and second ball drawn are black. We need to find $P(E \cap F)$.The total number of b

Vedclass · Accepted Answer

$\frac{3}{7}$

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(A) Let $E$ and $F$ denote respectively the events that the first and second ball drawn are black. We need to find $P(E \cap F)$.The total number of balls is $10 + 5 = 15$.The probability of drawing a black ball in the first draw is $P(E) = \frac{10}{15} = \frac{2}{3}$.After drawing one black ball without replacement,the number of black balls remaining is $9$ and the total number of balls remaining is $14$.The conditional probability of drawing a black ball in the second draw,given that the first ball was black,is $P(F|E) = \frac{9}{14}$.By the multiplication rule of probability,$P(E \cap F) = P(E) 	imes P(F|E)$.$P(E \cap F) = \frac{10}{15} 	imes \frac{9}{14} = \frac{2}{3} 	imes \frac{9}{14} = \frac{18}{42} = \frac{3}{7}$.

An urn contains $10$ black and $5$ white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

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