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.\(\frac{7}{9}\)
Answer: Option B

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

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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.    \(\frac{1}{221}\)
Answer: Option D

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.    \(\frac{5}{7}\)
Answer: Option C

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.    \(\frac{3}{25}\)
Answer: Option A

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.    \(\frac{1}{52}\)
Answer: Option C

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.    \(\frac{2}{77}\)
Answer: Option C

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.    \(\frac{13}{102}\)
Answer: Option D

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