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  • Need solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 19.

Need solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 19.

Answers (1)

Answer: \frac{10}{5}=\frac{5}{8}

Hint: Use combination method of n_{C_{r}}=\frac{n !}{(n-r) ! r !}

Given: O represents the event of getting two odd numbers

     S represents the event of getting their sum as an even number.

Solution:    

\begin{gathered} \mathrm{P}(\mathrm{S})=\left(4 \mathrm{C}_{2}+5 \mathrm{C}_{2}\right) / 9 \mathrm{C}_{2} \\ P(O \mid S)=\frac{n(O \cap S)}{n(S)} \\ =\frac{\frac{\mathrm{s}_{2}}{9} \mathrm{C}_{2}}{\frac{\left({ }^{4} \mathrm{C}_{2}+5_{2}\right)}{{ }^{9} \mathrm{C}_{2}}} \end{gathered}

=\frac{5 c_{2}}{\left(4 c_{2}+5_{c 2}\right)} \: \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\{\left(5_{C_{2}}\right)=\frac{5 \times 4 \times 3 \times 2 \times 1}{(5-2) !(2 \times 1)}=\frac{120}{3 \times 2 \times 1 \times 2}=10\right\}

\frac{10}{5}=\frac{5}{8}

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