Chapter 13: Probability

Given that a loaded die is thrown such that

P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3 and die is thrown two times. Also given that:

A = same number each time and

B = Total score is 10 or more.

So, P(A) = [P(1, 1) + P(2, 2) + P(3, 3) + P(4, 4) + P(5, 5) + P(6, 6)]

= P(1).P(1) + P(2).P(2) + P(3).P(3) + P(4).P(4) + P(5).P(5) + P(6).P(6)

= 0.2 x 0.2 + 0.2 x 0.2 + 0.1 x 0.1 + 0.3 x 0.3 + 0.1 x 0.1 + 0.1 x 0.1

= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20

Now, B = [(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)]

P(B) = [P(4).P(6) + P(6).P(4) + P(5).P(5) + P(5).P(6) + P(6).P(5) + P(6).P(6)]

= 0.3 x 0.1 + 0.1 x 0.3 + 0.1 x 0.1 + 0.1 x 0.1 + 0.1 x 0.1 + 0.1 x 0.1

= 0.03 + 0.03 + 0.01 + 0.01 + 0.01 + 0.01 = 0.10

A and B both events will be independent if

P(A ⋂ B) = P(A).P(B) …. (i)

And, here (A ⋂ B) = {(5, 5), (6, 6)}

So, P(A ⋂ B) = P(5, 5) + P(6, 6) = P(5).P(5) + P(6).P(6)

= 0.1 x 0.1 + 0.1 x 0.1 = 0.02

From equation (i) we get,

0.02 = 0.20 x 0.10

0.02 = 0.02

Therefore, A and B are independent events.

According to the solution of exercise 1, we have

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}and n(6) and n(S) = 6 x 6 = 36

So, P(A) = n(A)/n(S) = 6/36 = 1/6

And, B = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}; n(B) = 6 and n(S) = 36

So, P(B) = n(B)/n(S) = 6/36 = 1/6

Now, A ⋂ B = {(5, 5), (6, 6)}

So, P(A ⋂ B) = 2/36 = 1/18

Hence, if A and B are not independent, then

P(A ⋂ B) ≠ P(A).P(B)

1/18 ≠ 1/6 x 1/6 ⇒ 1/18 ≠ 1/36

Therefore, A and B are not independent events.

W.k.t, A ⋃ B denotes that atleast one of the events occurs and A ⋂ B denotes that two events occur simultaneously.

Therefore, the required answer is 1.1.

Let red marbles be presented with R and black marble with B. Also, let E be the event that at least one of the three marbles drawn be black when the first marble is red.

Now, the following three conditions are possible, if atleast one of the three marbles drawn be black and the first marble is red.

(i) E₁: 2^nd ball is black and 3^rd ball is red

(ii) E₂: 2^nd ball is black and 3^rd ball is also black

(iii) E₃: 2^nd ball is red and 3^rd ball is black

The probabilities of the above events are:

Therefore, the required probability is 25/56.

If two dice are thrown together, we have

n(S) = 36

Now, let’s consider:

E = A total of 4 = {(2, 2), (1, 3), (3, 1)} ⇒ n(E) = 3

F = A total of 9 or more

= {(3, 6), (6, 3), (5, 4), (4, 5), (5, 5), (4, 6), (6, 4), (5, 6), (6, 5), (6, 6)}

So, n(F) = 10

G = A total divisible by 5

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

So, n(G) = 7

It’s seen that (E ⋂ F) = Ø and (E ⋂ G) = Ø

And, (F ⋂ G) = {(4, 6), (6, 4), (5, 5)}

⇒ n(F ⋂ G) = 3 and (E ⋂ F⋂ G) = Ø

Hence, the probabilities if the events are:

As, P(F ⋂ G) ≠ P(F). P(G)

Therefore, there is no pair of independent events.

As the random variable X takes 0, 1, 2, 3, …, n is said to be binomial distribution having parameters n and p, if the probability is given by

P(X = r) = ⁿC_r p^r q^n-r, where q = 1 – p and r = 0, 1, 2, 3, …

Similarly, in case of tossing a coin 3 times, we have

n = 3 and X has the values 0, 1, 2, 3 with p = ½, q = ½.

Therefore, it is said that the experiment of tossing a coin three times have binomial distribution.

Given, P(A) = 1/2, P(B) = 1/3 and P(A Ç B) = 1/4.

P(A’) = 1 – 1/2 = 1/2, P(B’) = 1 – 1/3 = 2/3

Now,

Given, P(A) = 2/5, P(B) = 1/3 and P(C) = ½

P(A Ç C) = 1/5 and P(B Ç C) = ¼

So, P(C/B) = P(B Ç C)/ P(B) = (¼)/ (1/3) = ¾

P(A’ Ç C’) = 1 – P(A ⋃ C)

= 1 – [P(A) + P(C) – P(A Ç C)]

= 1 – [2/5 + ½ – 1/5] = 1 – 7/10 = 3/10

Therefore, the required probabilities are ¾ and 3/10.

Here, P(E₁) = p₁ and P(E₂) = p₂

Now, its clearly seen that either E₁ or E₂ occurs but not both.

For a probability distribution, we know that if P_i ≥ 0