Solution:

The total number of ways of selecting 4 tickets = 3⁴ = 81.

The favourable number of ways = sum of coefficients of x², x⁴,……. in (x + x² + x³)⁴

= sum of coefficients of x², x⁴,……. in x⁴ (1 + x + x²)⁴.

Let (1 + x + x²)⁴ = 1 + a₁x + a₂x² +…..+ a₈x⁸.

Then 3⁴ = 1 + a₁ + a₂ +…..+ a₈,

On putting x = 1

1 = 1 + a₁ + a₂ +…..+ a₈,

On putting x = −1

3⁴ + 1 = 2 [1 + a₁ + a₂ + a₄+ a₆ + a₈]

⇒ a₂ + a₄+ a₆ + a₈ = 41

Thus sum of the coefficients of x², x⁴,…… = 41

Hence the required probability = 41 / 81.

Solution:

x² + 2 (a + 4) x − 5a + 64 ≥ 0

If D ≤ 0, then (a + 4)² − (−5a + 64) < 0 Or

a² + 13a − 48 < 0 Or

(a + 16) (a − 3) < 0

⇒ −16 < a < 3 ⇔ −5 ≤ a ≤ 2

Then, the favourable cases are equal to the number of integers in the interval [−5, 2], i.e., 8.

The total number of cases is equal to the number of integers in the interval [−5, 30], i.e., 36.

Hence, the required probability is 8 / 36 = 2 / 9.

Solution:

We have the ratio of the ships A, B and C for arriving safely to be 2:5, 3:7 and 6:11 respectively. The probability of ship A for arriving safely = 2 / [2 + 5] = 2 / 7

Similarly, for B = 3 / [3 + 7] = 3 / 10 and for C = 6 / [6 + 11] = 6 / 17 Probability of all the ships for arriving safely = 2 / 7 × 3 / 10 × 6 / 17 = 18 / 595.

Solution:

Probability for white ball = 2 / 6 = 1 / 3

Probability for black ball = 4 / 6 = 2 / 3

Required probability = ⁵C₅ * (1 / 3)⁵ * (2 / 3)⁰ + ⁵C₄ * (1 / 3)⁴ * (2 / 3)

= (1/ 3)⁴ [(1 / 3) + [5] * [2 / 3]]

= 11 / 3⁵

= 11 / 243.

Solution:

Here P (X) = 0.3; P (Y) = 0.2

Now P (X ∪ Y) = P (X) + P (Y) − P (X ∩ Y)

Since these are independent events, so P (X ∩ Y) = P (X) * P (Y)

Thus required probability = 0.3 + 0.2 − 0.06 = 0.44

Solution:

The total number of digits in any number at the units place is 10.

Therefore, n (S) = 10

If the last digit is 1, 3, 5 or 7, then it is necessary that the last digit in each number must be 1, 3, 5 or 7.

Therefore, n (A) = 4

P (A) = 4 / 10 = 2 / 5

Hence, the required probability is (2 / 5)⁴ = 16 / 625.

Solution:

Let A be the event that the husband will be alive in 20 years and B be the event that the wife will be alive in 20 years. Clearly A and B are independent events.

P (A ∩ B) = P (A) * P (B)

Given P (A) = 3 / 5, P (B) = 2 / 3

The probability that at least of them will be alive in 20 years is

P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − P (A) P (B)

= 3 / 5 + 2 / 3 − [3 / 5] * [2 / 3]

= [9 + 10 − 6] / [15]

= 13 / 15

Solution:

Here P (without defect) = 8 / 10 = 4 / 5 = p

P (defected) = 2 / 10 = 1 / 5 = q and n = 2, r = 2

Hence required probability = ⁿC_r * p^r * q^n−r

= ²C₂* (45)² * (15)⁰

= 16 / 25

Solution:

Team totals must be 0, 1, 2, 39.

Let the teams be T1, T2,……., T40, so that T1 loses to T1 for i < j. In other words, this order uniquely determines the result of every game. There are 40! such orders and 780 games, so 2⁷⁸⁰ possible outcomes for the games.

Hence, the probability is 40! / 2⁷⁸⁰.

Solution:

Let p = Probability of getting tail = 1 / 2

q = Probability of getting head = 1 / 2

Also, p + q = 1 and n = 100

Required probability = P (X = 1) + P (X = 3) +….. + P (X = 99)

= ¹⁰⁰C₁ * p * q⁹⁹ + ¹⁰⁰C₃ * p³ * q⁹⁷ +……..+ ¹⁰⁰C₉₉ * p⁹⁹ * q¹

= [(p + q)¹⁰⁰ − (p − q)¹⁰⁰] / [2]

= 1 / 2.