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Probability (p-2) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Probability (p-2)2 Marks
Question
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random: passed at least one examination.
✓

Answer

Out of hundred students 1 student can be selected in ${ }^{100} \mathrm{C}_1=100$ ways.
$
\therefore \mathrm{n}(\mathrm{S})=100
$
Let $\mathrm{A}$ be the event that the student passed in the first examination.
Let $B$ be the event that student passed in second examination.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})=60, \mathrm{n}(\mathrm{B})=50 \text { and } \mathrm{n}(\mathrm{A} \cap \mathrm{B})=30 \\
& \therefore \quad \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{60}{100}=\frac{6}{10} \\
& \therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{50}{100}=\frac{5}{10} \\
& \\
& \therefore \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{30}{100}=\frac{3}{10} \\
&
\end{aligned}
$
$
\begin{aligned}: P(\text { student passed in at least one examination })=P(A \cup B) \\
& =P(A)+P(B)-P(A \cap B) \\
& =\frac{6}{10}+\frac{5}{10}-\frac{3}{10} \\
& =\frac{4}{5}
\end{aligned}
$

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