14.Probability
normal

A fair dice is thrown up to $20$ times. The probability that on the $10^{th}$ throw, the fourth six apears is :-

A

$\frac{{84 \times {5^6}}}{{{6^{10}}}}$

B

$\frac{{112 \times {5^6}}}{{{6^{10}}}}$

C

$\frac{{84 \times {5^6}}}{{{6^{20}}}}$

D

None

Solution

$^{9} \mathrm{C}_{3}\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{6} \times\left(\frac{1}{6}\right)$

$=\frac{84 \times 5^{6}}{6^{10}}$

Standard 11
Mathematics

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