Important Questions: Probability

# Important Questions: Probability | Mathematics (Maths) Class 12 - JEE PDF Download

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.

The document Important Questions: Probability | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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## Mathematics (Maths) Class 12

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## FAQs on Important Questions: Probability - Mathematics (Maths) Class 12 - JEE

 1. What is probability?
Ans. Probability is a measure of the likelihood that a particular event will occur. It is represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
 2. How is probability calculated?
Ans. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This can be expressed as a fraction, decimal, or percentage.
 3. What is the difference between theoretical and experimental probability?
Ans. Theoretical probability is based on mathematical calculations and assumptions, while experimental probability is determined through actual observations or experiments. Theoretical probability is often used to predict outcomes, while experimental probability is used to analyze real-world data.
 4. What is the addition rule of probability?
Ans. The addition rule of probability states that the probability of the union of two or more mutually exclusive events is the sum of their individual probabilities. In other words, if two events cannot occur simultaneously, the probability of either event happening is the sum of their individual probabilities.
 5. How does probability relate to statistics?
Ans. Probability is a fundamental concept in statistics. It provides the theoretical foundation for statistical analysis and inference. By understanding the probability of different outcomes, statisticians can make predictions, estimate parameters, and draw conclusions from data.

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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