Q1: Given that the events A and B are such that P(A) = 1/2, P (A ∪ B) = 3/5, and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent
Ans: Given, P(A) = 1/2 ,
P (A ∪ B) = 3/5
and P(B) = p.
(i) For Mutually Exclusive
Given that, the sets A and B are mutually exclusive.
Thus, they do not have any common elements
Therefore, P(A ∩ B) = 0
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Substitute the formulas in the above-given formula, we get
3/5 = (1/2) + p – 0
Simplify the expression, we get
(3/5) – (1/2) = p
(6 − 5)/10 = p
1/10 = p
Therefore, p = 1/10
Hence, the value of p is 1/10, if they are mutually exclusive.
(ii) For Independent events:
If the two events A & B are independent,
we can write it as P(A ∩ B) = P(A) P(B)
Substitute the values,
= (1/2) × p
= p/2
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Now, substitute the values in the formula,
(3/5) = (1/2)+ p – (p/2)
(3/2)– (1/2)= p – (p/2)
(6 − 5)/10 = p/2
1/10 = p/2
p = 2/10
P = 1/5
Thus, the value of p is 1/5, if they are independent.
Q2: 5 cards are drawn successively from a well-shuffled pack of 52 cards with replacement. Determine the probability that (i) all the five cards should be spades? (ii) only 3 cards should be spades? (iii) none of the cards is a spade?
Ans: Let us assume that X be the number of spade cards
Using the Bernoulli trial, X has a binomial distribution
P(X = x) = nCx qn-x px
Thus, the number of cards drawn, n = 5
Probability of getting spade card, p = 13/52 = 1/4
Thus the value of the q can be found using
q = 1 – p = 1 – (1/4)= 3/4
Now substitute the p and q values in the formula,
Hence, P(X = x) = 5Cx (3/4)5-x(1/4)x
(i) Probability of Getting all the spade cards:
P(all the five cards should be spade) = 5𝐶5 (1/4)5(3/4)0
= (1/4)5
= 1/1024
(ii) Probability of Getting only three spade cards:
P(only three cards should be spade) = 5𝐶3 (1/4)3(3/4)2
= (5!/3! 2!) × (9/1024)
= 45/ 512
(iii) Probability of Getting no spades:
P(none of the cards is a spade) = 5𝐶0(1/4)0(3/4)5
= (3/4)5
= 243/ 1024
Q3: A die is thrown twice and the sum of the numbers rising is noted to be 6. Calculate the is the conditional probability that the number 4 has arrived at least once?
Ans: If a dice is thrown twice, then the sample space obtained is:
S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(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)}
From the given data, it is needed to find the Probability that 4 has appeared at least once, given the sum of nos. is observed to be 6
Assume that, F: Addition of numbers is 6
and take E: 4 has appeared at least once
So, that, we need to find P(E|F)
Finding P (E):
The probability of getting 4 atleast once is:
E = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}
Thus , P(E) = 11/ 36
Finding P (F):
The probability to get the addition of numbers is 6 is:
F = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}
Thus, P(F) = 5/ 36
Also, E ∩ F = {(2,4), (4,2)}
P(E ∩ F) = 2/36
Thus, P(E|F) = (P(E ∩ F) ) / (P (F))
Now, subsbtitute the probability values obtained= (2/36)/ (5/36)
Hence, Required probability is 2/5.
Q4: The probability of solving the specific problem independently by the persons’ A and B are 1/2 and 1/3 respectively. In case, if both the persons try to solve the problem independently, then calculate the probability that the problem is solved.
Ans: Given that, the two events say A and B are independent if P(A ∩ B) = P(A). P(B)
From the given data, we can observe that P(A) = 1/2 & P(B) = 1/3
The probability that the problem is solved = Probability that person A solves the problem or the person B solves the Problem
This can be written as:
= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
If A and B are independent, then P(A ∩ B) = P(A). P(B)
Now, substitute the values,
= (1/2) × (1/3)
P(A ∩ B) = 1/6
Now, the probability of problem solved is written as
P(Problem is solved) = P(A) + P(B) – P(A ∩ B)
= (1/2) + (1/3) – (1/6)
= (3/6) + (2/6) – (1/6)
= 4/6
= 2/3
Hence, the probability of the problem solved is 2/3.
Q5: An fair die is thrown double times. Assume that the event A is “odd number on the first throw” and B the event “odd number on the second throw”. Compare the independence of the events A and B.
Ans: Let us consider two independent events A and B, then P(A ∩ B) = P(A). P(B)
when an unbiased die is thrown twice
S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(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)}
Let us describe two events as
A: odd number on the first throw
B: odd number on the second throw
To find P(A)
A = {(1, 1), (1, 2), (1, 3), …, (1, 6)
(3, 1), (3, 2), (3, 3), …, (3, 6)
(5, 1), (5, 2), (5, 3), …, (5, 6)}
Thus, P (A) = 18/36 = 1/2
To find P(B)
B = {(1, 1), (2, 1), (3, 1), …, (6, 1)
(1, 3), (2, 3), (3, 3), …, (6, 3)
(1, 5), (2, 5), (3, 5), …, (6, 5)}
Thus, P (B) = 18/36 = 1/2
A ∩ B = odd number on the first & second throw = { (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
So, P(A ∩ B) = 9/36 = 1/ 4
Now, P(A). P(B) = (1/2) × (1/2) = 1/4
As P(A ∩ B) = P(A). P(B),
Hence, the two events A and B are independent events.
