# A die is thrown 6 times. If getting an odd number is a success, what is the probability?

$\text {(i) of 5 successes? (ii) of at least 5 successes? (iii) of at most 5 successes?}$

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 toss the die 6 times. Getting an odd is considered a success. 6 tosses of the die are Bernoulli trial as they satisfy 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..
P (getting an odd number), $p = \large\frac{3}{6} = \frac{1}{2} \rightarrow$ $q = 1-p = 1 - \large\frac{1}{2} = \frac{1}{2}$
Let X be the number of times we get an odd number, i.e. a successful outcome.
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 $(q = 1 – p)$
$\Rightarrow P (X=x) = ^{6}C_x \times \large\frac{1}{2}^x $$\times \large\frac{1}{2}^{n-x} (i) Probability of 5 successes = P (X = 5). \Rightarrow P (X=5) =\; ^{6}C_5 \times \large\frac{1}{2}^5$$\times \large\frac{1}{2}^{6-5}$$=\;^{6}C_5 \times \large\frac{1}{2}^6 \Rightarrow P (X = 5) = 6 \times\large\frac{1}{64} = \frac{3}{32} (ii) Probability of at least 5 successes = P (X \geq 5) Given that there are only 6 tosses of the die. P (X \geq 5) = P(X = 5) + P(X = 6). We already know from above that P (X = 5) = \large\frac{3}{32} \Rightarrow P (X=6) =\; ^{6}C_6 \times \large\frac{1}{2}^6$$\times \large\frac{1}{2}^{6-6}$$=\;^{6}C_6 \times \large\frac{1}{2}^6$
$\Rightarrow P (X = 6) = \large\frac{1}{64}$
Therefore, P (X $\geq$ 5) = $\large \frac{6}{64} + \frac{1}{64} = \frac{7}{64}$
(iii) Probability of atmost 5 successes = P (X $\leq$ 5)
$\Rightarrow$ P (X $\leq$ 5) = 1 - P (X $\gt$ 5) = $1 - \large\frac{7}{64} = \frac{63}{64}$