Explain Solution R.D.Sharma Class 12 Chapter 30 Probability  Exercise 30.5 Question 33 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:\frac{1}{4}

Hint: You must know the rules of finding probability functions.

Given: Card is drawn from 52 cards, the outcome is noted , then it is replaced and reshuffled,

Another card is drawn.

Solution

We know that there are four suits club (C), spades(S), heart(H), diamond(D), each contain 13 cards.

P(both the cards are of same suit)

\begin{aligned} &=P\left[\left(C_{1} \cap C_{2}\right) \cup\left(S_{1} \cap S_{2}\right) \cup\left(H_{1} \cap H_{2}\right) \cup\left(D_{1} \cap D_{2}\right)\right] \\\\ &=P\left(C_{1} \cap C_{2}\right)+P\left(S_{1} \cap S_{2}\right)+P\left(H_{1} \cap H_{2}\right)+P\left(D_{1} \cap D_{2}\right) \\\\ &=P\left(C_{1}\right) \times P\left(C_{2}\right)+P\left(S_{1}\right) \times P\left(S_{2}\right)+P\left(H_{1}\right) \times P\left(H_{2}\right)+P\left(D_{1}\right) \times P\left(D_{2}\right) \\\\ &=\frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{13}{52} \\ &=4\left(\frac{1}{4} \cdot \frac{1}{4}\right) \\\\ &=\frac{1}{4} \end{aligned}

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