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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 19

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 19

Answers (1)

Answer:

 \frac{28}{45}

Hint:

 Use Baye’s theorem.

Given:

 In group of 400 people 160 smoke and non-vegetarians, 100 smoke and vegetarians. The probability of getting chest disease 35%, 20%, 10% respectively. A person chosen from group at found suffering. Find probability selected person smoke non vegetarian.

Solution:

Let us denote group of smoker and non-vegetarian as A.

B = smoker and non-vegetarian

C = no smoker and vegetarian

No of people in group A = 160;

P(A)=\frac{160}{400}

No of people in group B = 100;

P(B)=\frac{100}{400}

No of people in group C = 400 - 260 = 140;

\begin{aligned} &P(C)=\frac{140}{400}\\ &P(D_A)=35%\\ &P(D_B)=20%\\ &P(D_C)=10%\\ \end{aligned}

Probability of disease person from group A is

Using Baye’s theorem we get

\begin{aligned} &P(\text { smoker and vegetarian })=\frac{P(A)\times P(D_A)}{P(A)\times P(D_A)+P(B)\times P(D_B)+P(C)\times P(D_C)}\\ &=\frac{\frac{2}{5}\times 35%}{\frac{2}{5}\times 35%+\frac{1}{4}\times 20%+\frac{7}{20}\times 10%}\\ &=\frac{\frac{2}{5}\times 35}{\frac{2}{5}\times 35+\frac{1}{4}\times 20+\frac{7}{20}\times 10}\\ &=\frac{14}{14+5+\frac{7}{2}}\\ &=\frac{14}{\frac{19+7}{2}}\\ &=\frac{28}{38+7}\\ &=\frac{28}{45} \end{aligned}

Posted by

Gurleen Kaur

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