Theoretical and Experimental Probability

Illustrations:

Q1 : What is the probability of drawing 2 white and 5 green balls from a bag that contains 4 white and 10 green balls if 7 balls are drawn simultaneously at random.

Solution :

White balls 4

Green balls 10

Total balls 14

Let sample space be S and required event be E.

The number of ways 7 balls can be drawn from 14 balls in ¹⁴ C ₇

n(S) = ¹⁴C₇ = 3432

The number of ways of drawing 2 white and 5 green balls out of 4

white and 10 green balls is ⁴C₂X ¹⁰C₅ =6 X 252 = 1512

n(E) = 1512

P(E) = 1512/3432

= 0.44

Hence the probability of getting 2 white and 5 green balls from 4 white and 10 green balls is 0.44.

Q2 : From a normal pack of cards 4 cards are drawn. Find the probability that they are of different suits ?

Solution :

Total number of ways of drawing 4 cards from a pack of 52 cards is

n(S) = ⁵²C₄

= 270,725

Let E be the event of drawing 4 cards from different suits.

n(E) = ¹³c₁ X ¹³c₁ X ¹³c₁ X¹³c₁

= 13 X 13 X 13 X 13

=28561

P(E) = 28561 /270,725

= 0.105

Q3 : In a class there are 10 boys and 5 girls. A committee of 4 students are to be selected from the class. Find the probability for the committee to contain atleast 3 girls.

Solution :

Boys Girls Committee

(10) (5) (4)

1 3 ¹⁰C₁ X ⁵C₃ =10X10 =100

0 4 ¹⁰C₀ X ⁵C₄ =1 X 5 = 5

Four students can be selected from 10 +5 =15 students in ¹⁵C₄ ways

n(S) = ¹⁵C₄ =1365

Let E be the event of selecting atleast 3 girls to form a committee of 4 students.

n(E) = 100 + 5 = 105

P(E) = 105 /1365 = 0.077

Q4: If two cards are drawn from a well shuffled pack, find the probability that at least on the the two is hearts.

Solution :

Let S be the sample space associated with the drawing of two cards.

n(S) = ⁵²C₂ = 1326

Let E be the event of getting at least one of the two is hearts.

Then E^c be the event of getting none of them is hearts.

n(E^C )= ³⁹C₂ = 741

P(E^c ) = ³⁹C₂ / ⁵²C₂ = 741/1326

P(E) = 1 - P(E^c )

= 1 - ( ³⁹C₂ / ⁵²C_{2 )}

= 1 - 741/1326

= 0.441