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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise 30.5 Question 21 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 21 Maths Textbook Solution.

Answers (1)

Answer: \frac{2}{3}

Hint:  P=P\left ( A\cap B\cap \bar{C} \right )U\: P\left ( A\cap \bar{B}\cap{C} \right )U\: P\left ( \bar{A}\cap B\cap C \right )

Given: A can hit a target 3 times in 6 shots

B can hita target 2 times in 6 shots

C can hita target 4 times in 4 shots

Solution:

\begin{aligned} &P(\text { A hits the target })=\frac{3}{6}=\frac{1}{2} \\ &P(\mathrm{~B} \text { hits the target })=\frac{2}{6}=\frac{1}{3} \\ &P(\mathrm{C} \text { hits the target })=\frac{4}{4}=1 \end{aligned}

P (atleast 2 shots hit) = P(exactly 2 shots hit)+P(all 3 shots)

\begin{aligned} &=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(1-\frac{1}{2}\right)+\frac{1}{2} \times \frac{1}{3} \times 1 \\ &=\frac{1}{2}\left(\frac{2}{3}\right)+\frac{1}{3}\left(\frac{1}{2}\right)+\frac{1}{6} \\ &=\frac{2}{6}+\frac{1}{6}+\frac{1}{6} \\ &=\frac{2}{6}+\frac{1}{6}+\frac{1}{6} \\ &=\frac{2+1+1}{6}=\frac{4}{6}=\frac{2}{3} \end{aligned}

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