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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise 30.4 Question 24 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.4 Question 24 Maths Textbook Solution.

Answers (1)

Answer: No pair is independent

Hint: Probability=\frac{No\: of\: outcomes}{Total\: outcomes}

Given: Two dice are thrown together and the total score is noted. The event E,F and G are “a  total 4”, “a total of 9 or more” and “ a total divisible by 5”

Solution:

S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}

\begin{aligned} &n(S)=36 \\ &E=a \text { total of } 4 \\ &\quad=\{(1,3),(2,2),(3,1)\} \\ &\text { ie, } n(E)=3 \end{aligned}

F=a\: total\: of\: 9\: or\: more

       =\left\{\begin{array}{l} (3,6),(4,5),(4,6),(5,4),(5,5) \\ (5,6),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}

i.e,n(f)=10

G=a\: total\: divisible\: by\: 5

      =\left\{\begin{array}{l} (1,4),(2,3),(3,2),(4,1), \\ (4,6),(5,5),(6,4) \end{array}\right\}

i.e,n(G)=7

Now,

          \begin{aligned} &P(E)=\frac{3}{36}=\frac{1}{12} \\ &P(F)=\frac{10}{36}=\frac{5}{18} \\ &P(G)=\frac{7}{36} \end{aligned}

Also,

          \begin{aligned} &E \cap F=\varnothing, E \cap G=\varnothing \\ &F \cap G=\{(4,6),(5,5),(6,4)\} \\ &\text { ie, } n(F \cap G)=3 \\ \end{aligned}

\begin{aligned} &P(F) \times P(G)=\frac{5}{18} \times \frac{7}{36} \\ \end{aligned}

                                \begin{aligned} &=\frac{35}{648} \\ &\neq \frac{3}{36} \end{aligned}

i.e,P(F)\times P\left ( G \right )\neq P\left ( F\cap G \right )

So, no pair is independent.

Posted by

infoexpert21

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