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  • Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 6

Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 6

Answers (1)

Answer:

 \frac{193}{792}

Hint:

 To solve this first we find P(7 or 8) by head then P(7 & 8) by tails then required probability.

Given:

One coin – Head or Tail

Dice and 11 cards

Solution:

\text { Probability of an Head or Tail }= \left ( \frac{1}{2} \right )

Sample space of sum of the numbers of an unbiased dice when there will be head

\begin{aligned} &\text { Sample space }=\left\{\begin{matrix} 1+1; &2+1\dots &6+1 \\ 1+2; &2+2\dots &6+2 \\ 1+3; &2+3\dots &6+1 \\ \dots\dots &\dots\dots &\dots\dots \\ 1+6; &2+6\dots &6+6 \end{matrix}\right.\\ &n(S)=36\\ \end{aligned}

favourable outcome = 7 or 8 (Sum)                          
n  favour = 11 

\begin{aligned} &P (7\text { or } 8)=\frac{11}{36} \end{aligned}

Sample space of eleven cards when there will be tail,

Sample space = {2,3,4,5,6,7,8,9,10,11}

n(S)=11

favourable outcome = {7,8}, n  favour = 2

\begin{aligned} &P (7\text { or } 8)=\frac{2}{11}\\ &P (\text { 7 or 8 after getting head })=\frac{1}{2}\times \frac{11}{36}=\frac{11}{72}\\ & \text { Required prob (7 or 8) } =\frac{11}{72}+\frac{1}{11}\\ &=\frac{121+72}{72\times 11}\\ &=\frac{193}{792} \end{aligned}

Posted by

Gurleen Kaur

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