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

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 = ¹⁰C_{3 =}(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 × ⁶C₂ )

= 4 × (6 × 5) /  (2 × 1)   = 60

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

Solution 4            (Option D)

            Clearly, n(S) = ⁵²C_{2 =} (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 (E_{1) =} ²⁶C₂ = (26 ×25) / (2  × 1) = 325 ; n (E_{2) =} ⁴C_{2 =} (4 ×3) / (2  × 1) = 6

n ( E₁ ∩  E₂) = ²C_{2 =} 1

P (E₁  =   nE₁  / n (S)    =  325 / 1326            P (E₂) = nE2  / n (S)    =  6 / 1326  

             P (E₁ ∩  E₂)  = 1 / 1326

P (both red or both kings) = P (E₁  U  E₂ ) 

= P (E₁ )  + P (E₂ )  = 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

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