Question

Three cards are drawn from a pack of 52 cards. If the probability that they are of the same color is $\dfrac{k}{\lambda }$ then find $\lambda - k$.

Answer 1

Hint: Find the probability of the first card being red, then the second card being red, and then the third card being red. Multiply these to find the probability of drawing three red cards. Do the same for black cards. Add these probabilities to find the probability of drawing three cards of the same color.

Formula Used: \[{\text{Probability of an event = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]

Complete step by step Solution:
There are two cases where all three cards drawn are of the same color: they are all either red cards or all black cards.
Let the event of drawing 3 red cards be event L and the event of drawing 3 black cards be event M.
Let us first calculate $P(L)$
Probability of the first card being red, $P({L_1}) = \dfrac{{26}}{{52}}$
Probability of the second card being red provided the first card is red, $P({L_2}) = \dfrac{{25}}{{51}}$
Probability of the third card being red provided the first two cards are red, $P({L_3}) = \dfrac{{24}}{{50}}$
$P(L) = P({L_1}).P({L_2}).P({L_3})$
Therefore, $P(L) = \dfrac{{26}}{{52}}.\dfrac{{25}}{{51}}.\dfrac{{24}}{{50}} = \dfrac{2}{{17}}$
Similarly, $P(M) = \dfrac{2}{{17}}$
The probability of drawing three cards of the same color is the sum of the probability of drawing three red cards and the probability of drawing three black cards.
Therefore, the probability of drawing three cards of the same color is $P(L) + P(M) = \dfrac{4}{{17}}$
$k = 4,\lambda = 17$
Therefore, $\lambda - k = 13$

Note: The probability of the first card being red is $\dfrac{{26}}{{52}}$ because there are 26 red cards and 52 cards in total. The probability of the second card being red provided the first card is red is $\dfrac{{25}}{{51}}$ because there are 25 red cards left and 51 cards left in total. Using the same logic, the probability of the card being red is $\dfrac{{24}}{{50}}$.