Question

A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer 1

Hint:Use the fact that if A, B and C are independent events then $P\left( A\bigcap B\bigcap C \right)=P\left( A \right)P\left( B \right)P\left( C \right)$. Use the fact that if A, B and C are mutually exclusive events then $P\left( A\bigcup B\bigcup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)$. Use the fact that the probability of an event E is given by $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$, where n(E) is the number of elements in E and n(S) is the number of elements in the sample space. Use the fact that each toss of the die is independent of the other toss.

Complete step-by-step answer:
Let A: Getting an odd number on the first toss
B: Getting an odd number on the second toss
C: Getting an odd number on the third toss.
Here $P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{3}{6}=\dfrac{1}{2}$ and $P\left( A' \right)=1-\dfrac{1}{2}=\dfrac{1}{2}$
Hence, we have $P\left( A \right)=P\left( A' \right)=P\left( B \right)=P\left( B' \right)=P\left( C \right)=P\left( C' \right)=\dfrac{1}{2}$
Then E: The event of getting an odd number at least once is given by
$E=\left( A\bigcap B'\bigcap C' \right)\bigcup \left( A'\bigcap B\bigcap C' \right)\bigcup \left( A'\bigcap B'\bigcap C \right)\bigcup \left( A\bigcap B\bigcap C' \right)\bigcup \left( A\bigcap B'\bigcap C \right)\bigcup \left( A'\bigcap B\bigcap C \right)\bigcup \left( A\bigcap B\bigcap C \right)$
The above equation can be understood as follows:
The first three are for getting an odd number exactly once.
Then three are for getting an odd number exactly twice
The last one is for getting an odd number all the time.
Clearly, the union of those is equivalent to getting an odd number exactly once.
Hence, we have
$P\left( E \right)=P\left( \left( A\bigcap B'\bigcap C' \right)\bigcup \left( A'\bigcap B\bigcap C' \right)\bigcup \left( A'\bigcap B'\bigcap C \right)\bigcup \left( A\bigcap B\bigcap C' \right)\bigcup \left( A\bigcap B'\bigcap C \right)\bigcup \left( A'\bigcap B\bigcap C \right)\bigcup \left( A\bigcap B\bigcap C \right) \right)$
We know that if A, B and C are mutually exclusive events then $P\left( A\bigcup B\bigcup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)$.
Hence, we have
$P\left( E \right)=P\left( \left( A\bigcap B'\bigcap C' \right) \right)+P\left( \left( A'\bigcap B\bigcap C' \right) \right)+P\left( \left( A'\bigcap B'\bigcap C \right) \right)+P\left( \left( A\bigcap B\bigcap C' \right) \right)+P\left( \left( A\bigcap B'\bigcap C \right) \right)+P\left( \left( A'\bigcap B\bigcap C \right) \right)+P\left( \left( A\bigcap B\bigcap C \right) \right)$
Now, we know that if A, B and C are independent events then $P\left( A\bigcap B\bigcap C \right)=P\left( A \right)P\left( B \right)P\left( C \right)$.
Hence, we have
$P\left( \left( A\bigcap B'\bigcap C' \right) \right)=P\left( A \right)P\left( B' \right)P\left( C' \right)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{8}$
Similarly, we have
$\begin{align}
  & P\left( \left( A'\bigcap B\bigcap C' \right) \right)=\dfrac{1}{8} \\
 & P\left( \left( A'\bigcap B'\bigcap C \right) \right)=\dfrac{1}{8} \\
 & P\left( \left( A\bigcap B\bigcap C' \right) \right)=\dfrac{1}{8} \\
 & P\left( \left( A\bigcap B'\bigcap C \right) \right)=\dfrac{1}{8} \\
 & P\left( \left( A'\bigcap B\bigcap C \right) \right)=\dfrac{1}{8} \\
 & P\left( \left( A\bigcap B\bigcap C \right) \right)=\dfrac{1}{8} \\
\end{align}$
Hence, we have
$P\left( E \right)=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{7}{8}$
Hence the probability of getting an odd number at least once is equal to $\dfrac{7}{8}$.

Note: [1] Alternative Solution:
We have
E’: Getting an even number in all three tries
Hence, we have
$E'=A'\bigcap B'\bigcap C'$
Hence, we have
$P\left( E' \right)=P\left( A' \right)P\left( B' \right)P\left( C' \right)=\dfrac{1}{8}$
Hence we have
$1-P\left( E \right)=\dfrac{1}{8}\Rightarrow P\left( E \right)=1-\dfrac{1}{8}=\dfrac{7}{8}$
Hence the probability of getting an odd number at least once is equal to $\dfrac{7}{8}$.

[2] Alternative Solution:
The total number of elements in sample space $=6\times 6\times 6=216$
The number of ways of getting an event number in all three tries $=3\times 3\times 3=27$
Hence the number of elements in E is $=216-27=189$
Hence, we have $P\left( E \right)=\dfrac{189}{216}=\dfrac{7}{8}$