Before solving practice questions, go through Probability Basic Concepts

**Q1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3?**

a) 3/10 b) 3/20 c) 2/5

**Q2. In a class, 30% of the students offered English, 20% offered Hindi and 10% offered**

**both. If a student is selected at random, what is the probability that he has offered**

**English or Hindi?**

a) 2/5 b) 3/4 c) 3/5

d) 3/10 e) None of these

**Q3. A bag contains 6 white and 4 red balls. Three balls are drawn at random. What is the**

**probability that one ball is red and the other two are white?**

a) 1/2 b) 1/12 c) 3/10

d) 7/12 e) None of these

**Q4. Two cards are drawn from a pack of 52 cards. The probability that either both are red**

**or both are kings, is:**

a) 7/13 b) 3/26 c) 63/221

d) 55/221 e) None of these

**Q5. The probability that a card drawn from a pack of 52 cards will be a diamond or a king**

**is :**

a) 2/13 b) 4/13 c) 1/13

d) 1/52 e) None of these

###
**Answers**

**Solution 1 (**

**Option A)**

Here, S = [1, 2, 3, 4, …., 19, 20]

Let E = event of getting a multiple of 3 = [3, 6, 9, 12, 15, 18]

P (E) = n (E) / n (S) = 6 / 20 = 3 / 10

**Solution 2**

**(**

**Option A)**

P (E) = 30 / 100 = 3 / 10 , P (H) = 20 / 100 = 1 / 5 and P (E ∩ H) = 10 / 100 = 1 / 10

P (E or H) = P (E U H)

= P (E) + P (H) - P (E ∩ H)

= (3 / 10) + (1 / 5) - (1 / 10) = 4 / 10 = 2 / 5

**Solution 3**

**(**

**Option A)**

Let S be the sample space. Then,

n(S) = Number of ways of drawing 3 balls out of 10 =

^{10}C_{3 =}(10_{ }× 9 × 8) / (3 × 2 × 1)_{ }
Let E = event of drawing 1 red and 2 white balls

n(E) = Number of ways of drawing 1 red ball out of 4 and 2 white balls out of 6

= (

^{4}C_{1}×^{6}C_{2 })
= 4 × (6 × 5) / (2 × 1) = 60

P (E) = n(E) / n(S) = 60 / 20 = 1 / 2

**Solution 4**

**(**

**Option D)**

^{52}C

_{2 = }(52 × 51) / 2

Let E

_{1 }= event of getting both red cards,
E

_{2 }= event of getting both kings
Then, E

_{1}∩ E_{2 }= event of getting 2 kings of red cards.
n (E × 1) = 325 ; n (E × 1) = 6

_{1) = }^{26}C_{2 }= (26 ×25) / (2_{2) = }^{4}C_{2 =}(4 ×3) / (2_{}

n ( E

_{1}∩ E_{2}) =^{2}C_{2 = }1_{}

P (E = nE = 325 / 1326 P (E = 6 / 1326

_{1 }_{1 }/ n (S)_{ }_{2}) = nE2_{ }/ n (S)_{ }_{1 }∩ E

_{2}) = 1 / 1326

_{}

P (both red or both kings) = P (E

_{1 }U_{ }E_{2 })
= P (E

_{1 }) + P (E_{2 }) = P (E_{1}∩ E_{2})
= 325 / 1326 + 6 / 1326 - 1 / 1326

= 330 / 1326

= 55 / 221

**Solution 5**

**(**

**Option B)**

Here, n(S) = 52

There are 13 cards of diamond (including one king) and there are 3 more kings.

Let E = event of getting a diamond or a king.

Then, n(E) = (13 + 3) = 16

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