__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:**

- \(\frac{1}{15}\)
- \(\frac{25}{57}\)
- \(\frac{35}{256}\)
- \(\frac{1}{221}\)

**Answer: Option D**

Let S be the sample space.

Then, *n*(S) = ^{52}C_{2} = \(\frac{(52\times51)}{(2\times1)}\) = 1326.

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

So, *n*(E) = ^{4}C_{2} = \(\frac{(4\times3)}{(2\times1)}\) = 6.

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

__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:**

- \(\frac{21}{46}\)
- \(\frac{25}{117}\)
- \(\frac{1}{50}\)
- \(\frac{3}{25}\)

**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.

= ^{25}C_{3} `

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

= 2300.

*n*(E) = (^{10}C_{1} x ^{15}C_{2})

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

=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:**

- \(\frac{1}{13}\)
- \(\frac{2}{13}\)
- \(\frac{1}{26}\)
- \(\frac{1}{52}\)

**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:**

- \(\frac{1}{22}\)
- \(\frac{2}{22}\)
- \(\frac{2}{91}\)
- \(\frac{2}{77}\)

**Answer: Option C**

Let S be the sample space.

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

= ^{15}C_{3}

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

_{= 455.}

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

So, *n*(E) = ^{5}C_{3} = ^{5}C_{2} = \(\frac{(4\times5)}{(2\times1)}\) = 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:**

- \(\frac{3}{20}\)
- \(\frac{29}{34}\)
- \(\frac{47}{100}\)
- \(\frac{13}{102}\)

**Answer: Option D**

Let S be the sample space.

Then, *n*(S) = ^{52}C_{2} = \(\frac{(52\times51)}{(2\times1)}\) = 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

= (^{13}C_{1} x ^{13}C_{1})

= (13 x 13)

= 169.

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

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