Question

1. A card is drawn at random from an ordinary pack of 52 playing cards. Find the probability that the card is a seven.

2. An ordinary die is rolled once. Find the probability that

i. An even number occurs

ii. A number greater than 4 occurs

3. Examination results of 150 students showed that 95 students passed mathematics, 75

students passed economics and 135 students passed at least one of the above subjects. A

student is selected at random. What is the probability that the student

i. Passed both mathematics and economics?

ii. Failed both the subjects?

4. Three horses A, B, and C are in a race. A is twice as likely to win as B, and B is twice as

likely to win as C. What are their respective chances of winning?

5. The probability that a student passes history is 2/3 and that he passes economics is 4/9. If

the probability of passing at least one course is 4/5, what is the probability that he will

pass both courses?

Answer 1

1.) Total Cards = 52

Number of cards named '7' = 4

$P(X=4)=\dfrac{4}{52}=\dfrac{1}{13}$

2.) Total number of outcomes on rolling a die = 6

Outcomes = 1,2,3,4,5,6

(i) $P(even \ number)=\dfrac{3}{6}=\dfrac{1}{2}$

(ii) $P(greater \ than \ 4)=\dfrac{2}{6}=\dfrac{1}{3}$

3.) Given,

Total student = 150

Student who have passed mathematics M = 95

Students who have passed economics E = 75

Passed at least one subject $M\cup E=135$

(i) Number of students who passed both subjects $M\cap E=M+E-M\cup E$

$=95+75-135\\=35$

So, $P(passed \ both\ subject)=\dfrac{35}{150}=\dfrac{7}{30}$

(ii) Number of student who failed in both subject = 150 - 135 = 15

$P(\text{failed in both subjects})=\dfrac{15}{150}=\dfrac{1}{10}$

4.) Let the probability of winning of horse C is x.

then, probability of winning B is 2x and probability of winning A is 4x.

Also total probability of winning is 1.

So,

$P(A)+P(B)+P(C)=1\\ ⇒x+2x+4x=1\\⇒x=7$

​

Hence probability of winning of horses A,B, and C are $\dfrac{4}{7}\ ,\dfrac{2}{7}\ \ and\ \ \dfrac{1}{7}$ respectively.

5.)Given,

$P(H)=\dfrac{2}{3}\\\ \\P(E)=\dfrac{4}{9}\\\ \\P(H\cup E)=\dfrac{4}{5}$

We know that ,

$P(H\cup E)=P(H)+P(E)-P(H\cap E)$

So, Probability that student passes both the exam

$P(H\cap E)=P(H)+P(E)-P(H\cup E)$

$=\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{4}{5}\\\ \\=\dfrac{14}{15}$

Hence, $\boxed{P(H\cap E)=\dfrac{14}{45}}$