Question:

Two dice are tossed. The probability that the total score is a prime number is:

Options:
 A. $$\frac{1}{6}$$ B. $$\frac{5}{12}$$ C. $$\frac{1}{2}$$ D. $$\frac{7}{9}$$

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E  = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
(5, 2), (5, 6), (6, 1), (6, 5) }

So, n(E) = 15.

So, $$P(E) = \frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}.$$

Check All Probability Questions (MCQs)

## More Questions Related to Quantitative Aptitude > Probability :

Question 1.

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

Options:
1.    $$\frac{1}{15}$$
2.    $$\frac{25}{57}$$
3.    $$\frac{35}{256}$$
4.    $$\frac{1}{221}$$

Let S be the sample space.

Then, n(S) = 52C2 =  $$\frac{(52\times51)}{(2\times1)}$$   = 1326.

Let E = event of getting 2 kings out of 4.

So, n(E) = 4C2 = $$\frac{(4\times3)}{(2\times1)}$$  = 6.

So, $$P(E) = \frac{n(E)}{n(S)}=\frac{6}{1326}=\frac{1}{221}.$$

Question 2.

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Options:
1.    $$\frac{1}{10}$$
2.    $$\frac{2}{5}$$
3.    $$\frac{2}{7}$$
4.    $$\frac{5}{7}$$

P (getting a prize) =  $$\frac{10}{(10+25)}=\frac{10}{35}=\frac{2}{7}.$$

Question 3.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

Options:
1.    $$\frac{21}{46}$$
2.    $$\frac{25}{117}$$
3.    $$\frac{1}{50}$$
4.    $$\frac{3}{25}$$

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S)  = Number ways of selecting 3 students out of 25.

25C3 `

= $$\frac{(25\times24\times23)}{(3\times2\times1)}$$

= 2300.

n(E)  = (10C1 x 15C2)

$$\left[10\times\frac{(15\times14)}{(2\times1)}\right]$$

=1050.

So, $$P(E) = \frac{n(E)}{n(S)}=\frac{1050}{2300}=\frac{21}{46}.$$

Question 4.

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

Options:
1.    $$\frac{1}{13}$$
2.    $$\frac{2}{13}$$
3.    $$\frac{1}{26}$$
4.    $$\frac{1}{52}$$

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

So, $$P(E) = \frac{n(E)}{n(S)}=\frac{2}{52}=\frac{1}{26}.$$

Question 5.

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

Options:
1.    $$\frac{1}{22}$$
2.    $$\frac{2}{22}$$
3.    $$\frac{2}{91}$$
4.    $$\frac{2}{77}$$

Let S be the sample space.

Then, n(S)  = number of ways of drawing 3 balls out of 15

15C3

= $$\frac{(15\times14\times13)}{(3\times2\times1)}$$

= 455.

Let E = event of getting all the 3 red balls.

So,  n(E) = 5C3 = 5C2 =  $$\frac{(4\times5)}{(2\times1)}$$  = 10.

So, $$P(E) = \frac{n(E)}{n(S)}=\frac{10}{455}=\frac{2}{91}.$$

Question 6.

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

Options:
1.    $$\frac{3}{20}$$
2.    $$\frac{29}{34}$$
3.    $$\frac{47}{100}$$
4.    $$\frac{13}{102}$$

Let S be the sample space.

Then, n(S) = 52C2 =   $$\frac{(52\times51)}{(2\times1)}$$ = 1326.

Let E = event of getting 1 spade and 1 heart.

So, n(E)  = number of ways of choosing 1 spade out of 13 and 1 heart out of 13

= (13C1 x 13C1)

= (13 x 13)

= 169.

So, $$P(E) = \frac{n(E)}{n(S)}=\frac{169}{1326}=\frac{13}{102}.$$