Question

Suppose Komla throws a die repeatedly until he gets a six. What is the probability
that he needs to throw more than 10 times to get a six, to 3 decimal places?

Answer 1

Let X = the random variable denoting the number of throws Komla needs to get a six

Since, the probability of getting a six when a die is thrown is 1/6, then we have,

X ~ Geometric(p = 1/6)

The p.m.f. of X is given by,

P(X = x) = $\begin{cases} (1 - p)^{x-1}p, &\text{for }x = 1, 2, 3, .... \\ 0, &\text{otherwise} \end{cases}$

$\therefore$ The probability that Komla needs to throw more than 10 times to get a six

= P(X > 10)

= 1 - P(X $\leq$ 10) [since the total probability is 1]

= 1 - $\displaystyle\sum_{x=1}^{10}P(X=x)$

= 1 - $\displaystyle\sum_{x=1}^{10}(1 - p)^{x-1}p$

= 1 - $p[1 + (1 - p) + (1 - p)^2 + ... + (1 - p)^9]$

= 1 - $p\frac{1-(1-p)^{10}}{1-(1-p)}$

= 1 - $p\frac{1-(1-p)^{10}}{p}$

= 1 - 1 + (1 - p)¹⁰

= (1 - p)¹⁰ = (1 - 1/6)¹⁰ = (5/6)¹⁰ = 0.162 (rounded to 2 decimal places)

Answer: The probability that Komla needs to throw more than 10 times to get a six is 0.162.