When two are thrown then there are 6 × 6 exhaustive cases

∴ n = 36. Let A denote the event “total score of 7” when 2 dice are thrown then A

= [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)].

Thus there are 6 favourable cases.

∴ m = 6

By definition 

Total no. of ways of drawing 3 marbles

Total no. of ways of drawing marbles, which are of same colour

= ⁵C₃ + ⁴C₃ + ³C₃

= 10 + 4 + 1 = 15

∴ Probability of same colour

=  = ³⁄₄₄

∴ Probability of not same colour

= 1 – ³⁄₄₄ = ⁴¹⁄₄₄

n(S) = 

No. of selection of 3 oranges out of the total 12 oranges

= ¹²C₃ = 2 × 11 × 10 = 220.

No. of selection of 3 bad oranges out of the total 4 bad oranges = ⁴c₃ = 4

∴ n(E) = no. of desired selection of oranges

= 220 – 4 = 216

∴ P (E) = 

and 

⇒ p₁ > p₂

2 balls can be drawn in the following ways

1 red and 1 green or 2 red or 2 green

Required probability

= 

Since each ball can be put into any one of the three boxes. So, the total number of ways in which 12 balls can be put into three boxes is 3¹².

Out of 12 balls, 3 balls can be chosen in ¹²C₃ ways. Now, remaining 9 balls can be put in the remaining 2 boxes in 2⁹ ways. So, the total number or ways in which 3 balls are put in the first box and the remaining in other two boxes is ¹²C₃ × 2⁹.

Hence, required probability

Total number of balls = 12

Hence, required probability

No of ways of drawing 2 white balls from 5 white balls = ⁵C₂.

Also, No of ways of drawing 2 other from remaining 7 balls = ⁷C₂

There are 7 + 5 = 12 balls in the bag and the number of ways in which 4 balls can be drawn is ¹²C₄ and the number of ways of drawing 4 black balls (out of seven) is ⁷C₄.

Hence, P (4 black balls)

Thus the odds against the event ‘all black balls’ are

There are 5 + 7 = 12 balls in the bag and out of these two balls can be drawn in ¹²C₂ ways. There are 5 green balls, therefore, one green ball can be drawn in ⁵C₁ ways; similarly, one red ball can be drawn in ⁷C₁ ways so that the number of ways in which we can draw one green ball and the other red is ⁵C₁ × ⁷C₁.

Hence, P (one green and the other red)

If the drawn ball is neither red nor green, then it must be blue, which can be picked in ⁷C₁ = 7 ways. One ball can be picked from the total (8 + 7 + 6 = 21) in ²¹C₁ = 21 ways.

∴ Required probability