A bag contains 7 white, 5 black and 4 red balls. If two balls are draw

1.	A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that:(i) both the balls are white(ii) one ball is black and the other red(iii) both the balls are of the same colour
Answer» Given: A bag containing 7 white, 5 black and 4 red balls. By using the formula, P (E) = favourable outcomes / total possible outcomes Two balls are drawn at random, therefore Total possible outcomes are ¹⁶C₂ n (S) = 120 (i) Let E be the event of getting both white balls E = {(W) (W)} n (E) = ⁷C₂= 21 P (E) = n (E) / n (S) = 21 / 120 = 7/40 (ii) Let E be the event of getting one black and one red ball E = {(B) (R)} n (E) = ⁵C₁⁴C₁= 20 P (E) = n (E) / n (S) = 20 / 120 = 1/6 (iii) Let E be the event of getting both balls of same colour E = {(B) (B)} or {(W) (W)} or {(R) (R)} n (E) = ⁷C₂+ ⁵C₂+ ⁴C₂= 37 P (E) = n (E) / n (S) = 37/120

A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that:(i) both the balls are white(ii) one ball is black and the other red(iii) both the balls are of the same colour

Answer»

Given: A bag containing 7 white, 5 black and 4 red balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random, therefore

Total possible outcomes are ¹⁶C₂

n (S) = 120

(i) Let E be the event of getting both white balls

E = {(W) (W)}

n (E) = ⁷C₂= 21

P (E) = n (E) / n (S)

= 21 / 120

= 7/40

(ii) Let E be the event of getting one black and one red ball

E = {(B) (R)}

n (E) = ⁵C₁⁴C₁= 20

P (E) = n (E) / n (S)

= 20 / 120

= 1/6

(iii) Let E be the event of getting both balls of same colour

E = {(B) (B)} or {(W) (W)} or {(R) (R)}

n (E) = ⁷C₂+ ⁵C₂+ ⁴C₂= 37

P (E) = n (E) / n (S)

= 37/120

Discussion

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