Four candidates A, B, C, D have applied for the assignment

Question:

Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that

(a) C will be selected?

(b) A will not be selected?

Solution:

Given that A is twice as likely to be selected as B

i.e. P (A) = 2 P (B) ……1

and C is twice as likely to be selected as D

i.e. P (C) = 2 P (D) ……2

Now, B and C are given about the same chance

∴ P (B) = P (C) ……3

Since, sum of all probabilities = 1

∴ P (A) + P (B) + P (C) + P (D) = 1

⇒ P (A) + P (B) + P (B) + P (D) = 1 [from 3]

$\Rightarrow P(A)+\frac{P(A)}{2}+\frac{P(A)}{2}+\frac{P(C)}{2}=1_{[\text {from } 1 \& 2]}$

$\Rightarrow \frac{2 \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})}{2}=1_{[\text {from } 3]}$

$\Rightarrow 4 \mathrm{P}(\mathrm{A})+\frac{\mathrm{P}(\mathrm{A})}{2}=2$ [from 1]

$\Rightarrow \frac{8 P(A)+P(A)}{2}=2$

$\Rightarrow 9 P(A)=4$

$\Rightarrow P(A)=\frac{4}{9}$

(a) $P$ (C will be selected) $=P(C)$

$=P(B)[$ from 3$]$

$=\frac{\mathrm{P}(\mathrm{A})}{2}$ [from 1]

$=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$

(b) $\mathrm{P}$ (A will not be selected) $=\mathrm{P}\left(\mathrm{A}^{\prime}\right)$ By compliment rule

$=1-P(A)$

$=1-\frac{4}{9}$

= 5/9

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Comments

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Feb. 24, 2022, 7:56 p.m.
Plz find directly by putting eq in c. Instead of finding a first.
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