Find the probability of getting both red balls, when from a bag contai

1.	Find the probability of getting both red balls, when from a bag containing 5 red and 4 black balls, two balls are drawn, (i) with replacement (ii) without replacement
Answer» The bag contains 5 red and 4 black balls, i.e., 5 + 4 = 9 balls. (i) 2 balls can be drawn from 9 balls with replacement in ⁹C₁ X ⁹C₁ways. ∴ n(S) = ⁹C₁ X ⁹C₁= 9 × 9 = 81 Let event A: Balls drawn are red. 2 red balls can be drawn from 5 red balls with replacement in ⁵C₁X ⁵C₁ways. ∴ n(A) = ⁵C₁X ⁵C₁ = 5 × 5 = 25 ∴ P(A) = \(\frac {n(A)} {n (S)} = \frac {25}{81}\) (ii) 2 balls can be drawn from 9 balls without replacement in ⁹C₁X ⁸C₁ways. ∴ n(S) = ⁹C₁X ⁸C₁= 9 × 8 = 72 2 red balls can be drawn from 5 red balls without replacement in ⁵C₁X ⁴C₁ways. ∴ n(B) = ⁵C₁X ⁴C₁= 5 × 4 = 20 ∴ P(B) = \(\frac {n(B)} {n (S)} = \frac {20}{72} = \frac 5{18}\)

Find the probability of getting both red balls, when from a bag containing 5 red and 4 black balls, two balls are drawn, (i) with replacement (ii) without replacement

Answer»

The bag contains 5 red and 4 black balls, i.e., 5 + 4 = 9 balls.

(i) 2 balls can be drawn from 9 balls with replacement in ⁹C₁ X ⁹C₁ways.

∴ n(S) = ⁹C₁ X ⁹C₁= 9 × 9 = 81

Let event A: Balls drawn are red. 2 red balls can be drawn from 5 red balls with replacement in ⁵C₁X ⁵C₁ways.

∴ n(A) = ⁵C₁X ⁵C₁ = 5 × 5 = 25

∴ P(A) = \(\frac {n(A)} {n (S)} = \frac {25}{81}\)

(ii) 2 balls can be drawn from 9 balls without replacement in ⁹C₁X ⁸C₁ways.

∴ n(S) = ⁹C₁X ⁸C₁= 9 × 8 = 72

2 red balls can be drawn from 5 red balls without replacement in

⁵C₁X ⁴C₁ways.

∴ n(B) = ⁵C₁X ⁴C₁= 5 × 4 = 20

∴ P(B) = \(\frac {n(B)} {n (S)} = \frac {20}{72} = \frac 5{18}\)

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