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PROBABILITY — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsPROBABILITY3 Marks
Question
$3$ persons from medical profession and $5$ persons from engineering profession offer services at a social organization. $2$ persons are randomly selected from these persons with the purpose of forming a committee. Find the probability that both the persons selected belong to the same profession.
✓

Answer

There are in all $3+5=8$ persons. Hence, 2 persons can be selected in ${ }^{8} C_{2}=\frac{8 \times 7}{2 \times 1}=28$ ways.
Thus, the total number of mutually exclusive, exhaustive and equi-probable outcomes in the sample space is $n=28.$
Event that both the persons selected belong to medical profession $=A$
Even that both the persons selected belong to the engineering profession $=B$
Event that both the persons selected belong to the same profession $=A \cup B$
The two events $A$ and $B$ can not occur together that is $A \cap B=\phi$
Thus, the events $A$ and $B$ are mutually exclusive.
Hence, from the law of addition of probability, $P(A \cup B)=P(A)+P(B)$
For which we first find $P(A)$ and $P(B)$.
$A=$ Event that both the persons selected belong to medical profession.
The number of favourable outcomes of $A$ is $m={ }^{3} C_{2}=3$.
Probability of event $A \quad P(A)=\frac{m}{n}$
$=\frac{3}{28}$
$B=$ Event that both the persons selected belong to engineering profession.
The number of favourable outcomes of $B$ is $m={ }^{5} C_{2}=10$.
Probability of event $B \quad P(B)=\frac{m}{n}$
$=\frac{10}{28}$
Now,
$P(A \cup B) =P(A)+P(B)$
$ =\frac{3}{28}+\frac{10}{28}$
$ =\frac{3+10}{28}$
$ =\frac{13}{28}$
Required probability $=\frac{13}{28}$

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