PROBABILITY WORD PROBLEMS WITH SOLUTIONS

Problem 1 :

At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that (i) the first player wins a prize (ii) the second player wins a prize, if the first has won?

Solution :

Sample space  =  {1, 2, 3, 4, ..............1000}

n(S)  =  1000

Let "A" be the event of selecting a card that is a perfect square greater than 500.

A = { 23², 24², 25², 26², 27², 28², 29², 30², 31² }

n(A)  =  9

(i) the first player wins a prize

By picking one of the cards from A, he may win the prize

P(A)  =  n(A)/n(S)

P(A)  =  9/1000

(ii) the second player wins a prize, if the first has won?

"B" be the event of second player wins a prize.

n(B)  =  8, n(S)  =  1000 - 1  =  999 (the carded picked in (i) is not replaced)

P(B)  =  n(B)/n(S)

P(B)  =  8/999

Problem 2 :

A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.

Solution :

Total number of balls  =  blue balls + red balls

n(S)  =  12 + x 

(i) what is the probability that it will be a red ball? 

Let "A" be the event of selecting a red ball

n(A)  =  x

P(A)  =  n(A) / n(S)

P(A)  =  x / (12 + x)

(ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.

Number of red balls after placing 8 more red balls  =  x+8

P(B)  =  n(B)/n(S)

n(B)  =  (x + 8)

P(B)  =  (x + 8)/(12 + x + 8)

  =  (x + 8)/(20 + x)

P(B)  =  2P(A)

(x + 8)/(20 + x)  =  2x / (12 + x)

(x + 8)(12 + x)  =  2x (20 + x)

12x  + x² + 96 + 8x  =  40x + 2x²

2x²  - x² + 40x - 20x - 96  =  0

x² + 20x - 96  =  0

(x - 4) (x + 24)  =  0

x  =  4 and x = -24 (not admissible)

By applying the value of x in P(A), we get 

P(A)  =  4/(12 + 4)

  =  4/16

P(A)  =  1/4

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