Three faces of a fair die are yellow,two face | vedclass

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(B) The probabilities of getting yellow $(Y)$,red $(R)$,and blue $(B)$ in a single toss are: $P(Y) = \frac{3}{6} = \frac{1}{2}$,$P(R) = \frac{2}{6} =

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$\frac{1}{6}$

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(B) The probabilities of getting yellow $(Y)$,red $(R)$,and blue $(B)$ in a single toss are: $P(Y) = \frac{3}{6} = \frac{1}{2}$,$P(R) = \frac{2}{6} = \frac{1}{3}$,and $P(B) = \frac{1}{6}$.We want the probability that in $3$ tosses,we get one yellow,one red,and one blue face.The number of ways to arrange the sequence $(Y, R, B)$ is $3! = 6$.The probability of any one such sequence (e.g.,$Y, R, B$) is $P(Y) 	imes P(R) 	imes P(B) = \frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{6} = \frac{1}{36}$.Since there are $6$ such permutations,the total probability is $6 	imes \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$.

Three faces of a fair die are yellow,two faces are red and one face is blue. If the die is tossed $3$ times,then the probability that the colours yellow,red and blue appear is (need not be in that order).

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