There are 5 red balls ,4 yellow balls and 3 green balls in a basket.If 3 balls are randomly drawn then what is the probability that atleast 2 of them are green.?

option..a.1/11 b13/55 c3/11 d11/55 e 7/55 .?

Question

Answer

There are; 5 (red balls) + 4 (yellow balls) + 3 (green balls) = 12 balls.

Pr(at least 2 of them are green in colour) = The number of ways picking 3 balls at random and at least 2 of them are the green balls (which i will call “A”) devided by the number of ways If 3 balls are drawn at random (which i will call “B”).

Pr(at least 2 of them are green in colour) = A/B

A = (3C2 x 9C1) + (3C3 x 9C0) = 27 + 1 = 28

Explaination:

(3C2 x 9C1) = because at least 2 green balls out of 3 totals green balls and 1 balls out of total red and yellow balls.

(3C3 x 9C0) = because another choice is 3 green balls out of 3 totals green balls and none of yellow and red balls.

B = 12C3 = the number of ways If 3 balls are drawn at random . = 220

Pr(at least 2 of them are green in colour) = A/B

Pr(at least 2 of them are green in colour) = 28/220 = 7/55

Thus, option e) 7/55 is correct.

Answer

Solution:

Total number of balls = 5 + 4 + 3 = 12

Number of ways of selecting 3 balls out of 12 = C(12,3) = 220

Number of ways of selecting 2 green balls out of 3 = C(3,2) = 3

Number of ways of selecting 1 ball out of 9 (excluding green balls) = C(9,1) = 9

Required probability = (Number of ways of selecting 2 green balls and 1 ball out of 9) + (Number of ways of selecting 3 green balls)

= (C(3,2) * C(9,1)) / C(12,3) + C(3,3) / C(12,3)

= (3 * 9) / 220 + 1 / 220

= 28 / 220

= 7 / 55

Therefore, the probability that at least 2 balls are green is 7/55.

Answer: Option e. 7/55

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