Four candidates A, B, C, D have applied for the assignment to each a s

1.	Four candidates A, B, C, D have applied for the assignment to each 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 – (i) C will be selected? (ii) A will not be selected.
Answer» Let E₁, E₂, E₃, E₄ denote the events that the person A, B C and D is selected respectively. Also, let the probability of selecting D be ‘k’. Then, P(E₄) = k ∴ P(E₁) = 2P(E₂), P(E₂) P(E₃) = 2P(E₄) = 2k = 2 × 2k = 4k Since E₁, E₂, E₃, E₄ are mutually exclusive and exhaustive events so– P(E₁ ∪ E₂ ∪ E₃ ∪ E₄) = P(E₁) + P(E₂) + P(E₃) + P(E₄) ⇒ 1 = 4k + 2k + 2k + k ⇒ 9k = 1 ⇒ k = 1/9 (i) P(E₃) = 2k = \(\frac{2}{9}\) (ii) P(E₁’) = 1 – p(E₁) = 1 – 4k = 1 – 4(10) = 1 − \((\frac{1}{9})\) = 1 - \(\frac{4}{9}\) = \(\frac{5}{9}\)

Four candidates A, B, C, D have applied for the assignment to each 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 – (i) C will be selected? (ii) A will not be selected.

Answer»

Let E₁, E₂, E₃, E₄ denote the events that the person A, B C and D is selected respectively.

Also, let the probability of selecting D be ‘k’.

Then, P(E₄) = k

∴ P(E₁) = 2P(E₂), P(E₂) P(E₃) = 2P(E₄) = 2k

= 2 × 2k

= 4k

Since E₁, E₂, E₃, E₄ are mutually exclusive and exhaustive events so–

P(E₁ ∪ E₂ ∪ E₃ ∪ E₄) = P(E₁) + P(E₂) + P(E₃) + P(E₄)

⇒ 1 = 4k + 2k + 2k + k

⇒ 9k = 1

⇒ k = 1/9

(i) P(E₃) = 2k = \(\frac{2}{9}\)

(ii) P(E₁’) = 1 – p(E₁)

= 1 – 4k = 1 – 4(10)

= 1 − \((\frac{1}{9})\) = 1 - \(\frac{4}{9}\)

= \(\frac{5}{9}\)

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