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# Probability Problems with Detailed Solutions 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
d) ½                                  e) None of these

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

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 = 10C3 =(10  × 9  × 8) / (3 × 2  × 1)
= 120
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
= (4C1 × 6C)
= 4 × (6 × 5) /  (2 × 1)   = 60
P (E) = n(E) / n(S)   = 60 / 20   =  1 / 2

Solution 4            (Option D)

Clearly, n(S) = 52C2 = (52 × 51) / 2   = 1326
Let E1 = event of getting both red cards,
E2 = event of getting both kings
Then,  E1 ∩ E2 = event of getting 2 kings of red cards.
n (E1) = 26C2 = (26 ×25) / (2  × 1) = 325 ; n (E2) = 4C2 = (4 ×3) / (2  × 1) = 6
n ( E1 ∩  E2) = 2C2 = 1
P (E1  =   nE / n (S)    =  325 / 1326            P (E2) nE2  / n (S)    =  6 / 1326
P (E∩  E2)  = 1 / 1326
P (both red or both kings) = P (E1   E)

= P (E)  + P (E)  = P (E1 ∩  E2

=  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
P (E) = 16 / 52 = 4 / 13 