Question

A pack of cards contains 4 aces, 4 kings, 4 queens, and 4 jacks. Two cards are drawn at random from this pack without replacement. If the probability that at least one of them will be an ace is $\dfrac{\lambda }{k}$, then find $\lambda + k$.

Answer 1

Hint: Find the probability of the following three cases – both the cards are aces, the first card is an ace, the second card is not an ace, and the first card is not an ace, and the second card is an ace. Add the probabilities of all three cases to find the probability that at least one of them will be an ace.

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

Complete step by step Solution:
The total number of cars in this pack = 20
There are three cases in which at least one card is an ace:
a) Both cards are aces. Let this be event A.
There are 4 aces and 16 cards originally. So, the probability that the first card is an ace $ = \dfrac{4}{{16}}$.
Now there are 3 aces and 15 cards. So, the probability that the second card is an ace $ = \dfrac{3}{{15}}$.
Therefore, $P\left( A \right) = \dfrac{4}{{16}}.\dfrac{3}{{15}} = \dfrac{1}{{20}}$
b) The First card is an ace, and the second card is not an ace. Let this be event B.
There are 4 aces and 16 cards originally. So, the probability that the first card is an ace $ = \dfrac{4}{{16}}$.
Now there are 3 aces and 15 cards. So, the probability that the second card is not an ace $ = \dfrac{{12}}{{15}}$.
Therefore, $P\left( B \right) = \dfrac{4}{{16}}.\dfrac{{12}}{{15}} = \dfrac{1}{5}$
c) The First card is not an ace, and the second card is an ace. Let this be event C.
There are 4 aces and 16 cards originally. So, the probability that the first card is not an ace $ = \dfrac{{12}}{{16}}$.
Now there are 4 aces and 15 cards. So, the probability that the second card is an ace $ = \dfrac{4}{{15}}$.
Therefore, \[P\left( C \right) = \dfrac{{12}}{{16}}.\dfrac{4}{{15}} = \dfrac{1}{5}\]
Therefore, probability that at least one of the two cards will be an is $P\left( A \right) + P\left( B \right) + P\left( C \right) = \dfrac{1}{{20}} + \dfrac{1}{5} + \dfrac{1}{5} = \dfrac{9}{{20}}$.
Hence, $\lambda = 9$ and $k = 20$.
Therefore, $\lambda + k = 29$

Note: A common mistake in such a question is as follows- After considering the case where both the cards are aces, one might take a case where one card is an ace, and the other is not and add the two probabilities as $\dfrac{1}{{20}} + \dfrac{1}{5}$. This is wrong because we must be specific as to which one is an ace, and which one is not an ace.