3 Marks Question

Question 13 Marks

A couple has 2 children. Find the probability that both are boys, if it is known that (i) one of them is boy (ii) the older child is a boy.

Answer

Sample space $=\left\{B_1 B_2, B_1 G_2, G_1 B_2, G_1 G_2\right\}$ Where, $B_1$ and $G_1$ are the older boy and girl, respectively.
Let $\quad E_1=$ both the children are boys
$E_2=$ one of the children is a boy
$E_3=$ the older child is a boy.
(i) $P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
(ii) $P\left(\frac{E_1}{E_3}\right)=\frac{P\left(E_1 \cap E_3\right)}{P\left(E_3\right)}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$

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Question 23 Marks

Two thirds of the students in a class are boys and the rest girls. It is known that the probability of a girl getting a first class is 0.25 and that of a boy getting a first class is 0.28. Find the probability that a student chosen at random will get first class marks in the subject.

Answer

Let $E_1, E_2$ and $A$ be the events defined as follows:
$E_1=$ a boy is chosen from the class
$E_2=$ a girl is chosen from the class
$A=$ the student gets first class marks
Here, $P\left(E_1\right)=\frac{2}{3}, P\left(E_2\right)=\frac{1}{3}, P\left(\frac{A}{E_1}\right)=0.28$ and $P\left(\frac{A}{E_2}\right)=0.25$
Using the law of total probability, we obtain
$P(A)=P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A /E_2\right)$
$=\frac{2}{3} \times 0.28+\frac{1}{3} \times 0.25$
=0.27

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Question 33 Marks

If $A$ and $B$ are two events such that $P\left(\frac{A}{B}\right)=p$, $P(A)=p, P(B)=\frac{1}{3}$ and $P(A \cup B)=\frac{5}{9}$, then find the value of $p$.

Answer

Given, $p\left(\frac{A}{B}\right)=p, P(A)=p, P(B)=\frac{1}{3}$ and $P(A \cup B)$ $=\frac{5}{9}$
$\because \quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
$\Rightarrow \quad p=\frac{P(A \cap B)}{\frac{1}{3}}$
$\Rightarrow \quad P(A \cap B)=\frac{p}{3}$
We know that,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow \quad \frac{5}{9}=p+\frac{1}{3}-\frac{p}{3}$
$\Rightarrow \quad \frac{5}{9}-\frac{1}{3}=p-\frac{p}{3}$
$\Rightarrow \quad \frac{2}{9}=\frac{2 p}{3}$
$\Rightarrow \quad p=\frac{3}{9}=\frac{1}{3}$

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Question 43 Marks

If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$, then $P\left(\frac{A}{R}\right) \geq P(A)$ is true or not?

Answer

If $A \subset B$, then $A \cap B=A$
$\therefore \quad P(A \cap B)=P(A)$
Also, $\quad P(A) < P(B)$
Consider, $\quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}$$\quad$...(i)
It is known that $P(B) \leq 1$
$\therefore \quad \frac{1}{P(B)} \geq 1$
or $\quad \frac{P(A)}{P(B)} \geq P(A)$$\quad$...(ii)
From eqs. (i) and (ii), we get
$P\left(\frac{A}{B}\right) \geq P(A)$
Hence, the given statement is true.

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Applied Maths 3 Marks Question

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