Question

Find the probability of drawing a diamond card in each to two consecutive draws from a well- shuffled pack of cards, if the card drawn is not replaced after the first draw.

Answer 1

Hint: In this question, we will use the concept of multiplication theorems on probability which states that if A and B are two events associated with a random experiment, then P (A$ \cap $B) = P (A) P (B/A), if P (A) $ \ne $0. Here we will consider some events as according to the statements given in the question and then by using this theorem, we will find out the required probability.

Complete step-by-step solution -
Here, given that there’s a well shuffled pack of cards.
We know that in a pack of cards, there are a total of 52 cards, in which there are four suits, club, spade, diamond and heart, each having 13 cards.
Let A be the event of drawing a diamond card in the first draw and,
B be the event of drawing a diamond card in the second draw. Then,
\[ \Rightarrow \] P (A) = $\dfrac{{^{13}{C_1}}}{{^{52}{C_1}}} = \dfrac{{13}}{{52}} = \dfrac{1}{4}.$ [ $^n{C_1} = n$ ]
So now, after drawing a diamond card in the first draw 51 cards are left out of which 12 cards are diamond cards.
$\therefore $ P (B/A) = Probability of drawing a diamond card in the second draw when a diamond card has already been drawn in the first draw.
\[ \Rightarrow \] P (B/A) = $\dfrac{{^{12}{C_1}}}{{^{51}{C_1}}} = \dfrac{{12}}{{51}} = \dfrac{4}{{17}}.$
Hence, the required probability = P (A$ \cap $B) = P (A) P (B/A) = $\dfrac{1}{4} \times \dfrac{4}{{17}} = \dfrac{1}{{17}}$.
So this is the required probability.

Note: In this type of questions first we will make some events according to the statements given in the question and then we will find the probability of the event of drawing the first card and second card by removing the first card from the deck. After that we have used the multiplication theorem of probability to find out the required probability.