If three dice are thrown, then the probability that they show the numbers in A.P. is

\frac{1}{36}

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\frac{1}{18}

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\frac{2}{9}

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\frac{5}{18}

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Solution

The correct option is B $\frac{1}{18}$
A.P. with common diff one are $= 4 = (123) (234) (345) (456)$
A.P. with common diff $2$ are $= 2 = (135) (246)$
A.P. with common diff $0$ are $= 6$
$\Rightarrow P = 2 + 4 + 6 / 216 = 1 / 18$

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Similar questions

Three different dice are thrown three times. What is the probability that they show different numbers only two times?

ω

ω \neq 1

r_{1}, r_{2} a n d r_{3}

ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0,

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