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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: failed in both examinations.
✓

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}:\mathrm{P}(\text { student failed in both examinations })=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right) \\
& =\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime} \text {.....[De Morgan's law] } \\
& =1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\
& =1-\frac{4}{5} \\
& =\frac{1}{5}
\end{aligned}
$

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