Q.An urn contains 10 black and 5 white balls. Two balls are drawn one after another, and the first ball is not returned to the urn before the second is drawn. Assuming that each ball in the urn is equally likely to be drawn, find the probability that both balls drawn are black.

Multiply the probability of the first card being black by the (conditional) probability of the second card being black given the first was black.

$P(\text{1st black}) = \dfrac{26}{52}=\dfrac12$.

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Multiply the sequential probabilities of drawing black cards: $\tfrac{26}{52}\cdot\tfrac{25}{51}=\tfrac{25}{102}$, so (d). …

Multiply the probability of drawing a black card first by the probability of drawing a black card second (without replacement), since the events are dependent.

A standard deck has 26 black cards out of 52.

$P(\text{1st black}) = \dfrac{26}{52}$

After removing one black card (without replacement), 25 black cards remain out of 51 total:

$P(\text{2nd black} \mid \text{1st black}) = \dfrac{25}{51}$

$P(\text{both black}) = \dfrac{26}{52} \times \dfrac{25}{51} = \dfrac{650}{2652} = \dfrac{25}{102}$

This can also be checked with combinations: $\dfrac{\binom{26}{2}}{\binom{52}{2}} = \dfrac{325}{1326} = \dfrac{25}{102}$.

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$\dfrac{15}{56}$.

Bag has $5$ red $+3$ blue $=8$ balls; $3$ are drawn without replacement.

Total ways to draw $3$ balls: $\binom{8}{3}=56$.

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The key idea is to count the number of ways to pick one ball of each colour and divide by the total number of ways to draw any three balls. The probability is $\frac{12}{42} = \frac{2}{7}$.

Multiply the probabilities of drawing a queen first, then a queen from the remaining cards (dependent events, no replacement).

$P(\text{1st queen}) = \dfrac{4}{52}$. Given the first card was a queen, $P(\text{2nd queen}) = \dfrac{3}{51}$.

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