Browse Questions

# Find the probability of throwing at most 2 sixes in 6 throws of a single die.

$\begin{array}{1 1} \large(\frac{5}{6})^4 \frac{35}{18} \\ \large(\frac{5}{6})^5 \frac{35}{18} \\ \large(\frac{5}{6})^4 \frac{70}{18} \\\large(\frac{5}{6})^5 \frac{70}{18}\end{array}$

Toolbox:
• For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
• Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.
The experiment is to throw at most 2 sixes in $n=6$ throws of die.
It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.
Let X be the number of times we get 6 in six throws of a die. P (getting a 6) $= p = \large\frac{1}{6} \rightarrow $$q = 1 - p =\large \frac{5}{6} Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, P (X = x) = \large^{n}C_x. p^x.q^{n–x} where x = 0, 1, 2,...,n and (q = 1 – p) P (throwing at least 2 sixes) = P (X \leq 2) = P (X=0) + P (X=1) + P (X=2) P (X = 0) =^{6}C_0. \large\frac{1}{6}^0.\large\frac{5}{6}^{6–0} = \large(\frac{5}{6})^6 P (X = 1) =^{6}C_1. \large\frac{1}{6}^1.\large\frac{5}{6}^{6–1} = 6 \times \large\frac{1}{6} \times \large(\frac{5}{6})^5$$= \large(\frac{5}{6})^5$
$P (X = 2) =^{6}C_2. \large\frac{1}{6}^2\large\frac{5}{6}^{6–2}$$= 15 \large\frac{1}{6}^2\large(\frac{5}{6})^4$
Therefore $P (X \leq 2) = \large(\frac{5}{6})^6$ $+ \large(\frac{5}{6})^5$ $+ 15 \large\frac{1}{6}^2\large(\frac{5}{6})^4$
$P (X \leq 2) = \large(\frac{5}{6})^4 ( (\frac{5}{6})^2 + \frac{5}{6} + \frac{15}{36})$
$P (X \leq 2) = \large(\frac{5}{6})^4 \frac{25+30+15}{36}$
$P (X \leq 2) = \large(\frac{5}{6})^4 \frac{35}{18}$