JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Probability

Let

E_{1}, E_{2}, E_{3}

be three mutually exclusive events such that

P (E_{1}) = \frac{2 + 3 p}{6}, P (E_{2}) = \frac{2 - p}{8}

and

P (E_{3}) = \frac{1 - p}{2}

. If the maximum and minimum values of p are

p_{1}

and

p_{2}

, then

(p_{1} + p_{2})

is equal to :

Answer: (B)

Solution

For valid probabilities:

0 \leq P (E_{i}) \leq 1

for

i = 1, 2, 3

. From

P (E_{1})

0 \leq \frac{2 + 3 p}{6} \leq 1 \Rightarrow - \frac{2}{3} \leq p \leq \frac{4}{3}

. From

P (E_{2})

0 \leq \frac{2 - p}{8} \leq 1 \Rightarrow - 6 \leq p \leq 2

. From

P (E_{3})

0 \leq \frac{1 - p}{2} \leq 1 \Rightarrow - 1 \leq p \leq 1

. Combined:

- \frac{2}{3} \leq p \leq 1

. Since the events are mutually exclusive:

P (E_{1}) + P (E_{2}) + P (E_{3}) \leq 1

. This gives

\frac{2 + 3 p}{6} + \frac{2 - p}{8} + \frac{1 - p}{2} \leq 1

. Simplifying:

\frac{4 ( 2 + 3 p ) + 3 ( 2 - p ) + 12 ( 1 - p )}{24} \leq 1 \Rightarrow \frac{8 + 12 p + 6 - 3 p + 12 - 12 p}{24} \leq 1 \Rightarrow \frac{26 - 3 p}{24} \leq 1 \Rightarrow 26 - 3 p \leq 24 \Rightarrow p \geq \frac{2}{3}

. Thus

\frac{2}{3} \leq p \leq 1

, giving

p_{1} = 1, p_{2} = \frac{2}{3}

, and

p_{1} + p_{2} = \frac{5}{3}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option