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  • Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 11 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 11 maths textbook solution

Answers (1)

Answer: 

X 0 1 2 3 4
P\left ( X \right ) \frac{91}{323} \frac{455}{969} \frac{70}{323} \frac{10}{323} \frac{1}{969}


Hint: Use permutation and combination.

Given: Five defective mangoes are accidentally mixed with 15 good ones. 4 mangoes are drawn at random. Find the probability of the no. of defective mangoes.

Solution: x represents the number of defective mangoes drawn.

\therefore x can take values 0,1,2 or 3

\therefore there are total 20 mangoes (15 good +5 defective ) mangoes

n(s)= total possible ways of selecting 5 mangoes n(s)=20 C_{4}

n(s)= total possible ways of selecting 5 mangoes

n(s)=20 C_{4} P(x=0)=P( selecting no defective mango )

P(x=0)=\frac{15 C 4}{20 C 4}=\frac{15 \times 14 \times 13 \times 12}{20 \times 19 \times 18 \times 17}=\frac{91}{323}

P(x=1)=P( selecting 1 defective mango and 3 good mangoes )

P(x=1)=\frac{5 C _{1} \times 15 C _{3}}{20 C_{4}}=\frac{455}{969}

P(x=2)=P( selecting 2 defective mangoes and 2 good mangoes )

P(x=2)=\frac{5 C_{2}\times 15 C _{3}}{20 C _{4}}=\frac{70}{323}

P(x=3)=P( selecting 3 defective mangoes and 1 good mango )

=\frac{5 C _{3} \times 15 C_{1}}{20 C _{4}}=\frac{10}{323}

P(x=4)=P (selecting 4 defective mangoes and no good mango)

=\frac{5C_{4}}{20C_{4}}=\frac{1}{969}

Now, we have  

Following table represents the probability distribution of x:

X 0 1 2 3 4
P\left ( X \right ) \frac{91}{323} \frac{455}{969} \frac{70}{323} \frac{10}{323} \frac{1}{969}
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