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A fair dice is thrown twenty times.Find the probability that on he tenth throw the fourth six appears......... now the total number of possible outcomes is 3c9 after that what will we do?

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3 Answers 3

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If the tenth roll is the fourth $6$, you want the probability of rolling 3 times the $6$ in the first 9 rolls and then, the probability of rolling a $6$.

The $\binom93$ is just the way of distributing the 3 rolls of $6$ amongst the 9 rolls. What is the probability associated with each specific roll?

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Lets say we have to fill $10$ places with the outcomes of a fair dice which is rolled $10$ times.

Total no. of cases $=6^{10}$ because every place can be filled in $6$ ways, so it is $6*6*6...10\space times$.

We now have to count the no. of favorable cases i.e. $4^{th}$ six comes at $10^{th}$ place.

Fixing $6$ at $10^{th}$ place we have $9$ places to deal with.

We now have to place $3$ sixes at any of the first $9$ places.

$3$ places can be chosen out of $9$ places in $^9C_3$ ways.

For the rest $6$ places we can only have $1, 2, 3, 4\space or\space 5$.

Hence, Total no. of favorable cases $= ^9C_3\cdot 5^6$

$$\therefore Required\space Probability = \frac{^9C_3\cdot 5^6}{6^{10}}$$

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In order for the 4th six to appear on the 10th throw, you need 3 sixes to appear in the first 9 throws, and then you need the 10th throw to be a six.

Let $ X $ be the number of sixes obtained in the first 9 throws. Since the die is fair, the probability of getting a six is $ \frac{1}{6} $, which means that $ X \sim Binomial(n=9,p=\frac{1}{6}) $.

$$ P = P(X=3) \times P(\text{10th throw} = 6) = {{9}\choose{3}}\bigg(\frac{1}{6}\bigg)^3\bigg(1-\frac{1}{6}\bigg)^{9-3} \times \bigg(\frac{1}{6}\bigg) = {{9}\choose{3}}\bigg(\frac{1}{6}\bigg)^4\bigg(\frac{5}{6}\bigg)^{6} $$

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