Probability Questions and Answers

Total number of boys and girls = 50 + 18 = 68

Let B be the event that the selected person is a boy and G be the event that the selected person is a girl knows the cricket.

P (B) = \frac{50}{68} = \frac{25}{34} ÷ 2 (as half of boys knows cricket)

= \frac{25}{68}

P (G) = \frac{18}{68} = \frac{9}{34}  ÷ 2 (as half of girls knows cricket)

= \frac{9}{68}

Therefore, P (B or G) = P (B) + P (G)

= \frac{25}{68} + \frac{9}{68} = \frac{34}{68} = \frac{1}{2}

Given first dice = 6 means Sample Space Now = {(6, 1)(6,2) , (6, 3) (6, 4) (6, 5) (6, 6)}

Total of two dice > 8  =  ((6, 3) (6, 4) (6, 5) (6, 6))

Required Probability = \frac{4}{6}  = \frac{2}{3}

There are overall 18 players = 12 men and 6 women.

Therefore three women can be selected from 6 in ⁶C₃ ways

= \frac{6!}{3!3!}

= 20 ways.

Three men can be selected in ¹²C₃

=\frac{12!}{3!9!}

= \frac{12\times 11\times 10}{3\times 2\times 1}

= 220 ways.

Thus total selections for the favorable cases= 20 x 220 = 4400 ways.

The sample space will cover total cases of 6 people chosen from 12 men and 6 women = (12+6=18) people that is ¹⁸C₆

= \frac{18!}{6!12!}

= 18564

So probability = (favourable cases)/(Sample space)

= 4400/18564

= 1100/4641

: P (S) = Total lights

P(S) = 15

P(E) = Defective light

P(E) = 6

Therefore, the probability of selecting a defective light = P(S)/P(E)

= 6/15 = 2/5

Let us assume the green balls as z

Orange ball = 40

Total balls= z + 40

P (green balls) = z/z +40

P (orange balls) = 40/z+40

According to the question,

P (green balls) = 7/10 P (orange ball)

z/z+40 = 7/10 × 40/z+40

z = 7/10 × 40

z = 28

White balls = 8

Blue balls = 12

Total = 8 white + 12 blue = 20 balls

The probability of receiving a blue ball = 12/20 or 3/5

The probability of not receiving blue ball = 1 - Probability of receiving the blue ball

= 1 - 3/5

= 2/5

The sample space of two dice rolled together is:

S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }

Therefore, it is clear that there are no outcomes in which the total of two dices are 1. Thus the probability = P(E) = n(E) / n(S) = 0 / 36 = 0

The probability of having the sum as seven = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}, hence,

P(E) = n(E) / n(S) = 6 / 36

Except one outcome, which is { (6,6) } all other are equals to less than twelve.

P(E) = n(E) / n(S) = 35 / 36

Therefore, the probabilities are: 0, 6/36, 35/36

Total chocolates = 5 green + 3 pink + 4 purple = 12
The probability of taking out one green chocolate = 5/12
The probability of taking out another green chocolate = 5/12
The probability of taking out one purple chocolate = 4/12
The probability of taking out two green chocolates and one purple chocolate = 5/12 × 5/12 × 4/12

= 100/1728

= 25/432

Probability of drawing a heart (E) = 13/52 =1/4

The probability of drawing a face card (4 kings, 4 queens, 4 jacks) (F) = 12/52 = 3/13

The probability of either a face card or a heart required to be added separately. However, do not forget to subtract three face cards of heart in order not to count them two times.

The probability of drawing (face card and heart) = 3/52

Probability of selecting either a heart or a face P (E or F)

= P (E or F) = P (E) + P (F) - P (E and F)

= 1/4 + 3/13 - 3/52  = 22/52 = 11/26

Total men and women in group = 80 + 40 = 120

Let X be the event that the chosen person is  men and Y be the event that the chosen person is women knowing basketball.

P (X) = 80/120 = \frac{2}{3} \times \frac{1}{2}  (as half of men knows basketball) = 1/3

P (Y) = 40/120 = \frac{1}{3} \times \frac{1}{2} (as half of women knows basketball) = 1/6

Therefore, P (X or Y) = P (X) + P (Y)

= \frac{1}{3} + \frac{1}{6} = \frac{1}{2}