A box contains 4 blue, 3 red, 5 green and 6 yellow marbles. If three marbles are drawn at random, wh

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04 May 2022 Probability

A box contains 4 blue, 3 red, 5 green and 6 yellow marbles. If three marbles are drawn at random, what is the probability that at least one is green?

OPtion

1) 32/51
2) 143/408
3) 65/102
4) 265/408
5) 65/136
6) 65/408
7) 27/272
8) 5/408
9) 5/18
10) None of these

Solution

Blue=4, Red=3, Green=5, Yellow=6, Total=18
Number of ways of drawing 3 balls out of 18 = 18C3 = 816
Ways of selecting no marble Green = (18-5)C3 = 13C3 = 286
Probability of no marble selected is Green = 286/816
So, probability of selecting at least 1 Green = 1 - Probability of selecting no Green = 1 - (286/816) = 530/816 = 265/408

Alternate Solution:
Probability of at least 1 green out of 3 drawn = (1 Green & 2 Non Green) Or (2 Green & 1 Non Green) Or All 3 Green
= (5C1x13C2 + 5C2x13C1 + 5C3) / 18C3
= (390+130+10) / 816
= 530/816
= 265/408

Correct Option: 4 Best Solution (2)

Dr.Dipin Singh   2 years AGO

Reqd probability = 1-(13c3/18c3)= 265/408

Like? Yes (2) | No (1)  

RAMADURAI NATESAN   2 years AGO

p = getting ( 1 g ball + 2g balls + 3g balls )
= (5c1*13c2 + 5c2*13c1 +5c3) / 18c3 = 265/408

Like? Yes (1) | No (1)