तीन छात्रों S_1, S_2 और S_3 को हल करने के लिए | vedclass

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(A) $P(U) = 1 - P(S_1^{\prime} \cap S_2^{\prime} \cap S_3^{\prime}) = \frac{1}{2}$$\Rightarrow P(S_1^{\prime}) \cdot P(S_2^{\prime}) \cdot P(S_3^{\pri

Vedclass · Accepted Answer

$\frac{13}{36}$

Vedclass · Answer

(A) $P(U) = 1 - P(S_1^{\prime} \cap S_2^{\prime} \cap S_3^{\prime}) = \frac{1}{2}$$\Rightarrow P(S_1^{\prime}) \cdot P(S_2^{\prime}) \cdot P(S_3^{\prime}) = \frac{1}{2}$$\Rightarrow (1 - P(S_1))(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2} \dots (1)$$P(V) = \frac{P(S_1 \cap S_2^{\prime} \cap S_3^{\prime})}{P(S_2^{\prime} \cap S_3^{\prime})} = \frac{1}{10}$$\Rightarrow \frac{P(S_1) \cdot P(S_2^{\prime}) \cdot P(S_3^{\prime})}{P(S_2^{\prime}) \cdot P(S_3^{\prime})} = \frac{1}{10}$$\Rightarrow P(S_1) = \frac{1}{10}$$P(W) = P(S_2 \cap S_3^{\prime}) = P(S_2) \cdot P(S_3^{\prime}) = \frac{1}{12}$$\Rightarrow P(S_2)(1 - P(S_3)) = \frac{1}{12} \dots (2)$समीकरण $(1)$ से,$(1 - \frac{1}{10})(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2}$$\Rightarrow (1 - P(S_2))(1 - P(S_3)) = \frac{5}{9} \dots (3)$समीकरण $(2)$ को समीकरण $(3)$ से विभाजित करने पर: $\frac{P(S_2)}{1 - P(S_2)} = \frac{1}{12} 	imes \frac{9}{5} = \frac{3}{20}$$20 P(S_2) = 3 - 3 P(S_2) \Rightarrow 23 P(S_2) = 3 \Rightarrow P(S_2) = \frac{3}{23}$$P(S_2)$ का मान समीकरण $(2)$ में रखने पर: $\frac{3}{23}(1 - P(S_3)) = \frac{1}{12}$$1 - P(S_3) = \frac{23}{36} \Rightarrow P(S_3) = 1 - \frac{23}{36} = \frac{13}{36}$अतः,$P(T) = P(S_3) = \frac{13}{36}$.

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