Question

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?

Answer 1

Answer:

Step-by-step explanation:

Prob of getting at least one blue = 1 - Prob of getting NO blue balls.

Prob of getting NO Blue ball => Getting a ball from 6 Red or 2 Green or 3 Yellow.

No of ways of selecting 4 ball from 6R or 2 G or 3Y = (2+3+6)C1 = 11C4 = 330

Getting any 4 ball from bag = 15C4 = 1365

Prob of getting NO Blue ball = Possible Outcomes ÷ Total Outcomes = 330÷1365

Prob of getting at least one blue = 1 - (330÷ 1365) = 1035 ÷ 1365 = 69/91

Hence 69÷91 is the required probability.

Answer 2

Total number of marbles=6+4+2+3=15

Number of green=2

Number of yellow=3

p(both green or both yellow) =(2/15)(1/14) +(3/15)(2/14) =1/105+1/35=4/105

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