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Consider an unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice. If the dice is thrown thrice, the probability of obtaining red colour on top face of the dice at least twice is----

I am getting 0.75 can anyone confirm this ?
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$\binom{3}{2}*(\frac{1}{3})^{2}* \frac{2}{3} + \binom{3}{3}*(\frac{1}{3})^{3} = \frac{7}{27} = 0.259$
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Why can't we do like P[x>=2]=1-{P[x=0]+P[x=1]}  ?
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Best answer

0.2592

Unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice means $\frac{2}{6} = \frac{1}{3}$ is probability of each color comes on top.

Now ask Atleast two

For 2= $\binom{3}{2}(\frac{1}{3})^{2} \frac{2}{3}$ = $\frac{4}{9}$

For 3= $\binom{3}{3}(\frac{1}{3})^{3}$ = $\frac{1}{27}$

Total = $\frac{7}{27}= 0.252$

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@ Prashant.  Why can't we do like P[x>=2]=1-{P[x=0]+P[x=1]}  ?

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You can do it . answer is same . 1- 20/27
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P(Red>=2) → favourable/Total

Total  for 3 faces = $6^{3}/(2!*2!*2!)$ = 27 way

favourable, at least two red :

2 red 1 Green = $\binom{3}{1}$ = 3 way

2 red 1 Blue = $\binom{3}{1}$ = 3 way

3 red = 1 way

favourable  =3+3+1 = 7 way

P=7/27
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