Question

A fair die is rolled $10$ times.

What is the probability that an odd number $\left(1,3,\text{or}5\right)$ will occur less than $3$ times?

Answer 1

Step 1. Model the given situation as Bernoulli Trials:

Recall that for an experiment involving $n$ Bernoulli Trials, each having probability of success $p$, the probability of exactly $r$ successes is ${}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}$.

The probability of an odd number $\left(1,3,\text{or}5\right)$ appearing in a throw of a fair die is $\frac{3}{6}=\frac{1}{2}$.

The fair die is rolled $10$ times.

The situation given here is equivalent to $n=10$ Bernoulli Trials each having $p=\frac{1}{2}$ as the probability of success.

Here, success represents the event that an odd number $\left(1,3,\text{or}5\right)$ appears in a throw of a fair die.

Step 2. Determine the probability of no odd number appearing in $10$ throws of a fair die.

Determine the probability of no odd number appearing in $10$ throws of a fair die by substituting $n=10$, $p=\frac{1}{2}$, and $r=0$ in ${}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}$:

$\begin{array}{rcl}{}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}& =& {}^{10}C_0{\left(\frac{1}{2}\right)}^0{\left(1-\frac{1}{2}\right)}^{10-0}\\ & =& 1\times 1\times {\left(\frac{2-1}{2}\right)}^{10}\text{(\hspace{0.17em}Substituted}{}^{n}C_0=\frac{n!}{\left(n-0\right)!0!}=1\text{)}\\ & =& \frac{1}{2^{10}}\end{array}$

Thus, the probability of no odd number appearing in $10$ throws of a fair die is $\frac{1}{2^{10}}$.

Step 3. Determine the probability of exactly one odd number appearing in $10$ throws of a fair die:

Determine the probability of exactly one odd number appearing in $10$ throws of a fair die by substituting $n=10$, $p=\frac{1}{2}$, and $r=1$ in ${}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}$:

$\begin{array}{rcl}{}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}& =& {}^{10}C_1{\left(\frac{1}{2}\right)}^1{\left(1-\frac{1}{2}\right)}^{10-1}\\ & =& 10\times \frac{1}{2}\times {\left(\frac{2-1}{2}\right)}^9\text{(\hspace{0.17em}Substituted}{}^{n}C_1=\frac{n!}{\left(n-1\right)!1!}=n\text{)}\\ & =& 10\times \frac{1}{2}\times {\left(\frac{1}{2}\right)}^9\\ & =& 10\times \frac{1}{2^{10}}\\ & =& \frac{10}{2^{10}}\end{array}$
Thus, the probability of exactly one odd number appearing in $10$ throws of a fair die is $\frac{10}{2^{10}}$.

Step 4. Determine the probability of exactly two odd numbers appearing in $10$ throws of a fair die:

Determine the probability of exactly two odd numbers appearing in $10$ throws of a fair die by substituting $n=10$, $p=\frac{1}{2}$, and $r=2$ in ${}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}$:

$\begin{array}{rcl}{}^{n}C_r{p}^r{\left(1-p\right)}^{n-r}& =& {}^{10}C_2{\left(\frac{1}{2}\right)}^2{\left(1-\frac{1}{2}\right)}^{10-2}\\ & =& \frac{10!}{\left(10-2\right)!2!}\times \frac{1}{2^2}\times {\left(\frac{2-1}{2}\right)}^8\text{(\hspace{0.17em}Substituted}{}^{n}C_r=\frac{n!}{\left(n-r\right)!r!}\text{)}\\ & =& \frac{10!}{8!2!}\times \frac{1}{2^2}\times {\left(\frac{1}{2}\right)}^8\\ & =& \frac{10\times 9}{2}\times \frac{1}{2^{10}}\\ & =& 45\times \frac{1}{2^{10}}\\ & =& \frac{45}{2^{10}}\end{array}$
Thus, the probability of exactly two odd numbers appearing in $10$ throws of a fair die is $\frac{45}{2^{10}}$:

Step 5. Determine the probability of less than three odd numbers appearing in $10$ throws of a fair die:

Recall that $P(A\text{or}B\text{or}C)=P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$for mutually exclusive events $A$, $B$ and $C$.

The event of less than three odd numbers appearing in $10$ throws of a fair die includes all the three events for which probability is calculated above, namely, the event “no odd numbers appear”, the event “one odd number appears”, and the event “two odd numbers appear”.

Also, since any two of these three events cannot occur simultaneously, they are mutually exclusive.

Hence, the probability of any one of them happening is the sum of their individual probabilities.

Determine the probability of less than three odd numbers appearing in $10$ throws of a fair die by adding the probability of no odd number appearing in $10$ throws of a fair die $\frac{1}{2^{10}}$, the probability of exactly one odd number appearing in $10$ throws of a fair die $\frac{10}{2^{10}}$, and the probability of exactly two odd numbers appearing in $10$ throws of a fair die $\frac{45}{2^{10}}$ .

$\begin{array}{rcl}P(\text{less than three odd numbers appearing)}& =& P(\text{no odd numbers)}+P\left(\text{one odd number}\right)+P\left(\text{two odd numbers}\right)\\ & =& \frac{1}{2^{10}}+\frac{10}{2^{10}}+\frac{45}{2^{10}}\\ & =& \frac{1+10+45}{2^{10}}\\ & =& \frac{56}{2^{10}}\\ & =& \frac{7}{2^7}\\ & =& \frac{7}{128}\end{array}$

Hence, the probability of less than three odd numbers appearing in $10$ throws of a fair die is $\frac{\mathbf{7}}{\mathbf{128}}$.