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Question-1

Find the probability of getting a number less than 5 in a single throw of a die.

Solution:
Possible outcomes={1, 2, 3, 4, 5, 6}

Favorable outcomes = Getting a number less than 5 = {1, 2, 3, 4}

P(Getting a number < 5) =


 

Question-2

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing:
(i) an ace.
(ii) `2’ of spades.
(iii) `10’ of a black suit

Solution:
One card is drawn from a well-shuffled deck of 52 cards.

(i) An ace is drawn.

    Number of possible outcomes = 52

    Number of favorable outcomes = 4
    P( Drawing an ace) =

(ii) A`2’ of spades is drawn.

    Number of possible outcomes = 52

    Number of favorable outcomes = 1

    P( Drawing a`2’ of spades) =

(iii) A `10’ of a black suit is drawn.

     Number of possible outcomes = 52

     Number of favorable outcomes = 2

     P( Drawing a `10’ of a black suit) =


Question-3

17 cards numbered 1, 2, 3, ..., 16, 17 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is         
(i) odd.                       
(ii) a prime              
(iii) divisible by 3     
(iv) divisible by 3 and 2 both

Solution:
17 cards numbered 1, 2, 3, ..., 16, 17 are put in a box and mixed thoroughly. One person draws a card from the box.

(i) The number on the card is odd

    Number of possible outcomes = 17

    Number of favorable outcomes = 9 [i.e 1, 3, 5, 7, 9, 11, 13, 15, 17]

 P( Getting an odd number on the card ) =

(ii) The number on the card is a prime.

     Number of possible outcomes = 17

     Number of favorable outcomes = 7 [i.e 2, 3, 5, 7, 11, 13, 17]

  P( Getting a prime on the card ) =

(iii) The number on the card is divisible by 3

      Number of possible outcomes = 17

      Number of favorable outcomes = 5 [i.e 3, 6, 9, 12, 15]

     P( Getting a number divisible by 3 ) =

(iv) The number on the card is divisible by 3 and 2.

      Number of possible outcomes = 17

      Number of favorable outcomes = 2 [i.e 6, 12]

P(Getting a number divisible by 3 and 2 ) =



Question-4

A die is thrown once. Find the probability of getting
(i) an even number
(ii) a number greater than 3
(iii) a number between 3 and 6

Solution:
A die is thrown once.

(i) An even number

    Number of possible outcomes = 6 [i.e 1, 2, 3, 4, 5, 6]

    Number of favorable outcomes = 3 [i.e 2, 4, 6]

    P(Getting a even number) =

(ii) A number greater than 3

     Number of possible outcomes = 6 [i.e 1, 2, 3, 4, 5, 6]

     Number of favorable outcomes = 3 [i.e 4, 5, 6]

P(Getting a number greater than 3) =

(iii) A number between 3 and 6

     Number of possible outcomes = 6 [i.e 1, 2, 3, 4, 5, 6]

     Number of favorable outcomes = 2 [i.e 4, 5]

P( Getting a number between 3 and 6) =


Question-5

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.

Solution:
Let the number of blue balls in the bag be x.

The number of red balls in the bag is 5.

P(Drawing a blue ball) =

P( Drawing a red ball) =

P( Drawing a blue ball) = 2 × P(Drawing a red ball) = 2 ×

                                                                        i.e x = 10

The number of blue balls in the bag is 10.


Question-6

A bag contains 12 balls out of which x are white.
(i) If one ball is drawn at random, what is the probability that it will be a white ball?
(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (I case). Find x.

Solution:
A bag contains 12 balls out of which x are white.

(i) If one ball is drawn at random;
     P (Drawing a white ball ) =

(ii) 6 more white balls are put in the bag.
     Hence the number of balls in the bag = 12 + 6 = 18

     Number of white balls = x + 6
     P (Drawing a white ball )=

     P (Drawing a white ball in II case ) = 2 × P (Drawing a white ball in I case)

     = 2 ×

         i.e x = 3


Question-7

Five male and three female candidates are available for selection as on manager in a company. Find the probability that male is selected.

Solution:
Number of male candidates available = 5
Number of female candidates available = 3 Total number of candidates available = 8 Total outcomes = 8
Number of outcomes for a male getting selected = 5 Probability of a male getting selected =

Question-8

Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability of a number which is a perfect cube.

Solution:
Outcome of getting a perfect cube = {8, 27, 64} Number of outcomes of getting a perfect cube = 3 Probability of getting a perfect cube =

Question-9

A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning this bet?

Solution:
Total number of cards = 52
Total number of possible outcomes = 52
Number of favourable outcomes of getting a spade or an ace = 13 + 3 = 16
Probability of getting a spade or an ace if a card is drawn at random =
                                                                                                       =

Question-10

17 cards numbered 1, 2, 3, ..., 16, 17 are put into a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is a prime.

Solution:
Prime numbers = {2, 3, 5, 7, 11, 13, 17} Number of favourable outcomes = 7 Probability of getting a card with a prime number = .

Question-11

Tickets numbered 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?

Solution:
Number of ticket that bear numbers multiple of 3 or 7 is 3, 6, 7, 9, 12, 14, 15 and 18.
Hence probability = .

Question-12

Two dice are thrown. What is the probability of obtaining a total of 10 or 11?

Solution:
Let A be the event of getting a total of 10.
Let B be the event of getting a total of 11.

A = {(4, 6), (5, 5), (6,4)}
B = {(5, 6), (6, 5)}
n(A) = 3
n(B) = 2
Therefore P(A) =
         and P(B) =
Since A and B are mutually exclusive
P (A B) = P(A) + P(B) = +=
Therefore P(Obtaining a total of 10 or 11) = .


Question-13

Find the probability of choosing square numbers between 2 and 100.

Solution:
Sample space S = {2, 3, …, 100}
Therefore n(S) = 99
Let A denote the event of choosing a square number.
Then A = {4, 9, 16, 25, 36, 49, 64, 81, 100}
    n(A) = 9

Therefore P(A)=
 = .

Question-14

In a throw of a fair die, what is the probability of getting a number greater than 4?

Solution:
Let A denote the event of getting a number greater than 4 in a throw of a fair die.
Then A = {5, 6}
 n(A) = 2
Therefore P(A) = 
  =




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