प्रायिकता प्रश्नावली

Hint

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

Hint

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)

Hint

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

Hint

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

Hint

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

Hint

2 balls can be drawn in the following ways

1 red and 1 green or 2 red or 2 green

Required probability

=

Hint

and

⇒ p₁ > p₂

Hint

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) =

Hint

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

Hint

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

Hint

Let W stand for the winning of a game and L for losing it. Then there are 4 mutually exclusive possibilities

(i) W, W, W
(ii) W, W, L, W
(iii) W, L, W, W
(iv) L, W, W, W.

[Note that case (i) includes both the cases whether he losses or wins the fourth game.]

By the given conditions of the question, the probabilities for (i), (ii), (iii) and (iv) respectively are

Hence the required probability

=

[Since the probability of winning the game if previous game was also won is 2/(1+2) = 2/3 and the probability of winning the game if previous game was a loss is 1/(1+2) = 1/3].

Hint

Probability that only husband is selected

Probability that only wife is selected

=

∴ Probability that only one of them is selected

Hint

Total of seven can be obtained in the following ways

1, 1, 1, 4 in = 4 ways [there are four objects, three repeated]

Similarly,

1, 1,2, 3 in = 12 ways

1, 2,2, 2 in = 4 ways

Hence, required probability

[Since exhaustive no. of cases = 6 × 6 × 6 × 6 = 6⁴]

Hint

The number of ways of getting the different number 1, 2, ….., 6 in six dice = 6!.

Total number of ways = 6⁶

Hence, required probability

=

Hint

The faulty machines can be identified in two tests only if both the tested machines are either all defective or all non-defective. See the following tree diagram.

(Here D is for Defective & ND is for Non Defective)

Required Probability

=

Since the probability that first machine is defective (or non-defective) is ²⁄₄ and the probability that second machine is also defective (or non – defective) is ⅓ as 1 defective machine remains in total three machines.

Hint

The probability that the person hits the target = 0.3

∴ The probability that he does not hit the target in a trial

= 1 – 0.3 = 0.7

∴ The probability that he does not hit the target in any of the ten trials = (0.7)¹⁰

∴ Probability that he hits the target

= Probability that at least one of the trials succeeds

= 1 – (0.7)¹⁰.

Hint

The probability that A cannot solve the problem

The probability that B cannot solve the problem

The probability that both A and B cannot solve the problem

∴ The probability that at least one of A and B can solve the problem

∴ The probability that the problem is solved = ¹¹⁄₁₂

Hint

Seven people can seat themselves at a round table in 6! ways. The number of ways in which two distinguished persons will be next to each other = 2 (5) !, Hence, the required probability

Hint

The condition implies that the last digit in both the integers should be 0, 1, 5 or 6 and the probability

[ Since the squares of numbers ending in 0 or 1 or 5 or 6 also 0 or 1 or 5 or 6 respectively]

Hint

Exhaustive number of cases = 12

Favourable cases = ¹²C₂ (2⁶ – 2)

∴ Probability