Chapter 13: Probability

Let’s assume X to be the random variable denoting a bulb to be defective.

Here, n = 10, p = 1/50, q = 1 – 1/50 = 49/50

We know that, P(X = r) = ⁿC_r p^r q^{n – r}

(i) None of the bulbs is defective, i.e., r = 0

P(x = 0) = ¹⁰C₀ (1/50)⁰(49/50)^{10 – 0} = (49/50)¹⁰

(ii) Exactly two bulbs are defective

So, P(x = 2) = ¹⁰C₂ (1/50)²(49/50)^{10 – 2}

= 45.49⁸/50¹⁰ = 45 x (1/50)¹⁰ x 49⁸

(iii) More than 8 bulbs work properly

We can say that less than 2 bulbs are defective

P(x < 2) = P(x = 0) + P(x = 1)

Let’s consider

E₁ = Event that the coin is fair

E₂ = Event that the coin is 2 headed

And H = Event that the tossed coin gets head.

Now,

P(E₁) = ½, P(E₂) = ½, P(H/E₁) = ½, P(H/E₂) = 1

Using Baye’s Theorem, we get

Therefore, the required probability is 1/3.

Let’s assume

E₁ = The event that a person selected is of blood group O

E₂ = The event that the people selected is of other group

And H = The event that selected person is left handed

Now,

P(E₁) = 0.30 and P(E₂) = 0.70

Further, P(H/E₁) = 0.06 and P(H/E₂) = 0.10

Using Baye’s Theorem, we have

Therefore, the required probability is 9/44.

Given, S = {1, 2, 3, …, n}

So, P(r ≤ p/s ≤ p) = P(P ⋂ S)/ P(S)

= p – 1/n x n/(n – 1)

= (p – 1)/(n – 1)

Therefore, the required probability is (p – 1)/(n – 1).

Let X be the random variable scores when a die is thrown twice.

X = 1, 2, 3, 4, 5, 6

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

So,

P(X = 1) = 1/6 . 1/6 = 1/36

P(X = 2) = 1/6 . 1/6 + 1/6 . 1/6 + 1/6 . 1/6 = 3/36

P(X = 3) = 1/6 . 1/6 + 1/6 . 1/6 + 1/6 . 1/6 + 1/6 . 1/6 + 1/6 . 1/6 = 5/36

Similarly,

P(X = 4) = 7/36; P(X = 5) = 9/36 and P(X = 6) = 11/36

Hence, the required distribution is

Therefore, the required mean = 161/36.

Given, X = 0, 1, 2 and P(X = 0) = P (X = 1) = p

Let P(X) at X = 2 is x

⇒ p + p + x = 1 ⇒ x = 1 – 2p

Now, we have the following distribution:

So,

E(X) = 0.p + 1.p + 2(1 – 2p) = p + 2 – 4p = 2 – 3p

And, E(X²) = 0.p + 1.p + 4(1 – 2p) = p + 4 – 8p = 4 – 7p

Also, given E(X) = E(X²)

2 – 3p = 4 – 7p

4p = 2

p = ½

Therefore, the required value of p is ½.

We know that,

Variance(X) = E(X²) – [E(X)]²

Therefore, the required variance is 665/324.

Let’s take A1 to be the event of getting a total of 6

A₁ = {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)}

And, B₁ be the event of getting a total of 7

B₁ = {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)}

Let P(A₁) is the probability, if A wins in a throw = 5/36

And P(B₁) is the probability, if B wins in a throw = 1/6

Therefore, the required probability of wining A in his third throw

Given, two events A and B are independent such that A = {(x, y) : x + y = 11} B = {(x, y) : x ¹ 5}

Now,

A = {(5, 6), (6, 5)}

B = {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (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)}

n(A) = 2, n(B) = 30 and n(A ⋂ B) = 1

So, P(A) = 2/36 = 1/18 and P(B) = 30/36 = 5/6

Now,

P(A).P(B) = 1/18. 5/6 = 5/108 and P(A ⋂ B) 1/36

As P(A). P(B) ≠ P(A ⋂ B)

Therefore, events A and B are not independent.

Let’s consider A to be the event of having m white and n black balls

E₁ = First ball drawn of white colour

E₂ = First ball drawn of black colour

E₃ = Second ball drawn of white colour

Therefore, the probability if drawing a white ball does not depend upon k.

Long Answer (L.A.)