Let E, and E, be two independent events. Let P(E1) denotes the probability of the occurrence of the event E Further. let E'1 and E'2 denote the complements of E1 and E2, respectively. If PE1′∩E2=215 and PE1∩E2′ =16, then PE1 is

PE1=m⇒PE11=1−mPE11∩E2=PE11⋅PE2=215⇒(1−m)PE2=215⇒PE2=215(1−m) and PE1∩E21=16⇒PE1⋅PE21=16⇒m1−215(1−m)=16⇒6m[15(1−m)−2]=15(1−m)⇒2m(13−15m)=5−5m⇒30m2−31m+5=0⇒m=56 or 15∴PE1=56 or 15.

Let E, and E, be two independent events. Let P(E1) denotes the probability of the occurrence of the event E Further. let E'1 and E'2 denote the complements of E1 and E2, respectively. If PE1′∩E2=215 and PE1∩E2′ =16, then PE1 is

PE1=m⇒PE11=1−mPE11∩E2=PE11⋅PE2=215⇒(1−m)PE2=215⇒PE2=215(1−m) and PE1∩E21=16⇒PE1⋅PE21=16⇒m1−215(1−m)=16⇒6m[15(1−m)−2]=15(1−m)⇒2m(13−15m)=5−5m⇒30m2−31m+5=0⇒m=56 or 15∴PE1=56 or 15.

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Let E, and E, be two independent events. Let P(E₁) denotes the probability of the occurrence of the event E Further. let E'₁ and E'₂ denote the complements of E₁ and E₂, respectively. If $P (E_{1}^{'} \cap E_{2}) = \frac{2}{15}$ and $P (E_{1} \cap E_{2}^{'})$ $= \frac{1}{6}, then P (E_{1})$ is

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Detailed Solution

$P (E_{1}) = m \Rightarrow P (E_{1}^{1}) = 1 - m$
$P (E_{1}^{1} \cap E_{2}) = P (E_{1}^{1}) \cdot P (E_{2}) = \frac{2}{15}$
$\begin{array}{l} \Rightarrow (1 - m) P (E_{2}) = \frac{2}{15} \Rightarrow P (E_{2}) = \frac{2}{15 (1 - m)} \\ and P (E_{1} \cap E_{2}^{1}) = \frac{1}{6} \Rightarrow P (E_{1}) \cdot P (E_{2}^{1}) = \frac{1}{6} \\ \Rightarrow m (1 - \frac{2}{15 (1 - m)}) = \frac{1}{6} \\ \Rightarrow 6 m [15 (1 - m) - 2] = 15 (1 - m) \\ \Rightarrow 2 m (13 - 15 m) = 5 - 5 m \\ \Rightarrow 30 m^{2} - 31 m + 5 = 0 \\ \Rightarrow m = \frac{5}{6} or \frac{1}{5} \\ ∴ P (E_{1}) = \frac{5}{6} or \frac{1}{5} . \end{array}$