Let E1, E2, E3 be three mutually exclusive events such that P(E1) = <math><mrow><mfrac><mrow><mn>2</mn><mo>+</mo><mn>3</mn><mi>p</mi></mrow><mrow><mn>6</mn></mrow></mfrac></mrow></math> , P(E3) = <math><mrow><mfrac><mrow><mn>2</mn><mo>−</mo><mi>p</mi></mrow><mrow><mn>8</mn></mrow></mfrac></mrow></math> and P(E3) = <math><mrow><mfrac><mrow><mn>1</mn><mo>−</mo><mi>p</mi></mrow><mrow><mn>2</mn></mrow></mfrac><mo>.</mo></mrow></math> If the maximum and minimum values of p are p1 and p2, then (p1 + p2) is equal to:

0≤2+3p6≤1⇒p∈ [−23, 43], 0≤2−p8≤1⇒p∈ [−6, 2] and 0≤1−p2≤1⇒p∈ [−1, 1]0<P (E1)+P (E2)+P (E3)≤10≤1312−p8≤1⇒p∈ [23, 263] Taking intersection to all p∈ [23, 1]∴p1+p2=53

Let E1, E2, E3 be three mutually exclusive events such that P(E1) =&nbsp;<math><mrow><mfrac><mrow><mn>2</mn><mo>+</mo><mn>3</mn><mi>p</mi></mrow><mrow><mn>6</mn></mrow></mfrac></mrow></math> , P(E3) =&nbsp;<math><mrow><mfrac><mrow><mn>2</mn><mo>−</mo><mi>p</mi></mrow><mrow><mn>8</mn></mrow></mfrac></mrow></math>&nbsp;and P(E3) =&nbsp;<math><mrow><mfrac><mrow><mn>1</mn><mo>−</mo><mi>p</mi></mrow><mrow><mn>2</mn></mrow></mfrac><mo>.</mo></mrow></math> If the maximum and minimum values of p are p1 and p2, then (p1 + p2) is equal to:

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Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let E₁, E₂, E₃ be three mutually exclusive events such that P(E₁) = $\frac{2 + 3 p}{6}$ , P(E₃) = $\frac{2 - p}{8}$ and P(E3) = $\frac{1 - p}{2} .$ If the maximum and minimum values of p are p₁ and p₂, then (p₁ + p₂) is equal to:

Option 1 -

$\frac{2}{3}$

Option 2 -

$\frac{5}{3}$

Option 3 -

$\frac{5}{4}$

Option 4 -

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$0 \leq \frac{2 + 3 p}{6} \leq 1 \Rightarrow p \in [- \frac{2}{3}, \frac{4}{3}], 0 \leq \frac{2 - p}{8} \leq 1 \Rightarrow p \in [- 6, 2] a n d 0 \leq \frac{1 - p}{2} \leq 1 \Rightarrow p \in [- 1, 1]$

$0 < P (E_{1}) + P (E_{2}) + P (E_{3}) \leq 10 \leq \frac{13}{12} - \frac{p}{8} \leq 1 \Rightarrow p \in [\frac{2}{3}, \frac{26}{3}]$ Taking intersection to all $p \in [\frac{2}{3}, 1]$

$∴ p_{1} + p_{2} = \frac{5}{3}$

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Let E₁, E₂, E₃ be three mutually exclusive events such that P(E₁) = $\frac{2 + 3 p}{6}$ , P(E₃) = $\frac{2 - p}{8}$ and P(E3) = $\frac{1 - p}{2} .$ If the maximum and minimum values of p are p₁ and p₂, then (p₁ + p₂) is equal to:

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Let E1, E2, E3 be three mutually exclusive events such that P(E1) = 2+3p6 , P(E3) = 2−p8 and P(E3) = 1−p2. If the maximum and minimum values of p are p1 and p2, then (p1 + p2) is equal to:

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Let E₁, E₂, E₃ be three mutually exclusive events such that P(E₁) = $\frac{2 + 3 p}{6}$ , P(E₃) = $\frac{2 - p}{8}$ and P(E3) = $\frac{1 - p}{2} .$ If the maximum and minimum values of p are p₁ and p₂, then (p₁ + p₂) is equal to: