4 Marks Questions

Question 14 Marks

A young man visits a hospital for medical check up. The probability that he has lungs problem is 0.55 , heart problem is 0.29 and either lungs or heart problem is 0.57 . What is the probability that he has
(i) both type of problems : lungs as well as heart ?
(ii) lungs problem but not heart problem ?
Out of $1 0 0 0$ persons, how many are expected to have both type of problems?

Answer

Let ' $L$ ' and ' $H$ ' be the events that the young man has lungs problem and heart problem respectively.
$\begin{aligned} \therefore P(L)=0.55, P(H) & =0.29, P(L \cup H)=0.57 \\ P(L \cap H) & =?\end{aligned}$
(i) We know that, $P(L \cup H)=P(L)+P(H)-P(L \cap H)$
$\begin{aligned} \Rightarrow \quad P(L \cap H) & =P(L)+P(H)-P(L \cup H) \\ & =0.55+0.29-0.57 \\ & =0.27\end{aligned}$
$\therefore$ The probability that he has both the problems is 0.27
(ii)
$
\begin{aligned}
P\left(L \cap H^{\prime}\right) & =P(L)-P(L \cap H) \\
& =0.55-0.27 \\
& =0.28 .
\end{aligned}
$
$\therefore$ The probability that he has lungs problem but not heart problem is 0.28 .
$\therefore$ Out of 1000 persons, the number of persons having both the problems $=0.27 \times 1000=270 . $

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Question 24 Marks

In a single throw of two dice, determine the probability of getting a total of 7 or 9 .

Answer

Let $S$ be the sample space. Then, $n(S)=6^2=36$.
Let ' $A$ ' and ' $B$ ' be the events of getting a total of 7 and 9 respectively.
$\begin{array}{r}\therefore \quad A=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}\end{array}$
$\Rightarrow \quad n(A)=6$
$B=\{(3,6),(4,5),(5,4),(6,3)\}$
$\Rightarrow \quad n(B)=4$
$A \cap B=\phi \Rightarrow n(A \cap B)=0$
$\begin{aligned} \therefore \quad P(A) =\frac{n(A)}{n(S)}=\frac{6}{36} \\ P(B)=\frac{n(B)}{n(S)}=\frac{4}{36}\end{aligned}$
and $\quad P(A \cap B)=\frac{n(A \cap B)}{n(S)}=0$
We know that,
$\begin{aligned} P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{6}{36}+\frac{4}{36}-0 \\ & =\frac{10}{36}=\frac{5}{18}\end{aligned}$
$\therefore$ The probability of getting a total of 7 or 9 is $\frac{5}{18}$.

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Question 34 Marks

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is either a king or spade.

Answer

Let ' $S$ ' be the sample space.
So, $n(S)=52$ Also, let ' $A$ ' and ' $B$ ' be the event of getting a king and spade respectively.$
\therefore n(A)=4, n(B)=13 \text { and } n(A \cap B)=1
$
[ $\because$ there is only one card which is king of spade]
$\begin{array}{l} \Rightarrow P(A)=\frac{n(A)}{n(S)}=\frac{4}{52}=\frac{1}{13} \\ P(B)=\frac{n(B)}{n(S)}=\frac{13}{52}=\frac{1}{4}\end{array}$
$P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{52}$
We know that
$
P(A \cup B)=P(A)+P(B)-P(A \cap B)
$
$\begin{array}{l}=\frac{1}{13}+\frac{1}{4}-\frac{1}{52} \\ =\frac{17}{52}-\frac{1}{52} \\ =\frac{16}{52}\end{array}$
$=\frac{4}{13}$
$\therefore$ Required probability that the card is a king or a spade is $\frac{4}{13}$.

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