Let A and B be two events such that P ( A ∪ B ¯ ) = 1 6 , P ( A ∩ B ) = 1 4 and P ( A ¯ ) = 1 4 . Then the events A and B are

Independent but not equally likely

Equally likely but not independent

Mutually inclusive and dependent

A problem in Mathematics is given to three students whose chances of solving it are 1 3 , 1 4 , and 1 5 What is the probability that the problem is solved?

Let , A 1 denote the event of that first student solves the problem. A 2 denote the event that second student solves the problem. A 3 denote the event that third student solves the problem. Given P A 1 = 1 3 , P A 2 = 1 4 and P A 3 = 1 5 Problem will be solved if at least one of them solves it. P (at least one of them solves the problem) = 1 - P (No one solving the problem) = 1 - P A ¯ 1 ∩ A ¯ 2 ∩ A ¯ 3 = 1 - P A ¯ 1 · P A ¯ 2 · P A ¯ 3 Since if A 1 , A 2 , A 3 are independent then, A ¯ 1 , A ¯ 2 , A ¯ 3 are also independent. = 1 - 1 - p A 1 1 - P A 2 1 - P A 3 = 1 - 1 - 1 3 1 - 1 4 1 - 1 5 = 1 - 2 3 3 4 4 5 = 1 - 2 5 = 3 5 Therefore, Probability that the problem will be solved is 3 5 .

Given P ( A ) = 0 . 4 and P ( A ∪ B ) = 0 . 7 . Find P ( B ) if A and B are independent events.

Given that, P ( A ) = 0 . 4 P ( A ∪ B ) = 0 . 7 and A and B are independent events ∴ P ( A ∩ B ) = P ( A ) . P ( B ) We know that P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) ⇒ P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ) · P ( B ) ⇒ 0 . 7 = 0 . 4 + P ( B ) - 0 . 4 · P ( B ) ⇒ 0 . 7 - 0 . 4 = P ( B ) ( 1 - 0 . 4 ) ⇒ 0 . 3 = 0 . 6 × P ( B ) ⇒ P ( B ) = 0 . 3 0 . 6 = 3 6 = 1 2 = 0 . 5

Of the students in a college, it is known that 60 % reside in hostel and 40 % are day scholars (not residing in hostel). Previous year results report that 30 % of all students who reside in hostel attain A grade and 20 % of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?

Consider E 1 and E 2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A . ∴ P E 1 = 60 % = 60 100 = 0.6 P E 2 = 40 % = 40 100 = 0.4 P A | E 1 = P (student getting an A grade is a hostler) = 30 % = 0.3 P A | E 2 = P (student getting an A grade is a day scholar) = 20 % = 0.2 So, the probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P E 1 A . By using Bayes’ theorem, we get: P ( E 1 | A ) = P E 1 × P A E 1 P E 1 × P A E 1 + P E 2 × P A E 2 = 0.6 × 0.3 0.6 × 0.3 + 0.4 × 0.2 = 0.18 0.26 = 9 13 Hence, the required probability is 9 13

A family has two children. What is the probability that both children are boys, given that at least one of them is a boy?

Given, a family has two children. Let A be the event that both children are boys B be the event that at least one of them is a boy. Then, A = B B and B = B B , B G , G B A ∩ B = B B From the given data, P A = 1 4 P B = 3 4 and P A ∩ B = 1 4 The required probability is P A B = P A ∩ B P B = 1 / 4 3 / 4 P A / B = 1 3 Therefore, the required probability is 1 3 .

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white?

Given that, a bag contains 5 white, 7 red and 4 black balls. Here, total number of balls in the bag = 5 + 7 + 4 = 16 Probability of getting a white ball = 5 16 Probability of not getting white ball = 11 16 Four balls are drawn one by one with replacement Therefore, Probability that none is white = 11 16 × 11 16 × 11 16 × 11 16 = 11 16 4 = 14641 65536 Hence, the probability of that none is white = 14641 65536 .

The persons A , B and C shoot to hit a target. Their probabilities of hitting the target are 5 6 , 4 5 and 3 4 respectively. Find the probability that at least one person hits the target.

Here, the probabilities of hitting the target by three persons A , B and C are 5 6 , 4 5 and 3 4 respectively. i.e. P ( A ) = 5 6 , P ( B ) = 4 5 , P ( C ) = 3 4 . So, the required probability i.e. at least one person hits the target is = 1 - P A ¯ ∩ B ¯ ∩ C ¯ = 1 - P A ¯ · P B ¯ · P C ¯ = 1 − 1 − 5 6 1 − 4 5 1 − 3 4 = 1 − 1 6 × 1 5 × 1 4 = 1 − 1 120 = 119 120 = 0.991

Bag A contains 4 white balls and 3 black balls, while Bag B contains 3 white balls and 5 black balls. Two balls are drawn from Bag A and placed in Bag B . Then, what is the probability of drawing a white ball from Bag B ?

Let E 1 , E 2 , E 3 and A be the events defined E 1 = Two white balls is transferred from bag A to bag B E 2 = Two black balls is transferred from bag A to bag B E 3 = One white and one black balls is transferred from bag A to bag B A = One ball drawn from the bag is white. P ( E 1 ) = 4 C 2 7 C 2 = 4 × 3 2 × 1 7 × 6 2 × 1 = 2 7 [Using C r n = n ! r ! ( n - r ) ! ] P ( E 2 ) = 3 C 2 7 C 2 = 3 × 2 2 × 1 7 × 6 2 × 1 = 1 7 P ( E 3 ) = 4 C 1 × 3 C 2 7 C 2 = 4 × 3 7 × 6 = 4 7 When E 1 has occurred, two white balls has already been transferred from bag A to bag B contains 5 white balls and 5 black balls. P A E 1 = 5 10 When E 2 has occurred two black balls has already been transferred from bag A to bag B contains 3 white balls and 7 black balls. P A E 2 = 3 10 When E 3 has occurred one black ball and one white balls has already been transferred from bag A to bag B contain 4 white balls and 6 black balls. P A E 3 = 4 10 By using law of total probability, P ( A ) = P ( E 1 ) P A E 1 + P ( E 2 ) P A E 2 + P ( E 3 ) P A E 3 = 2 7 5 10 + 1 7 3 10 + 4 7 4 10 = 10 70 + 3 70 + 16 70 = 29 70 Hence, the required probability is 29 70 .

If A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 , then find P B A .

Given that A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 To find P B A : We know that P B A = P A ∩ B P A ⇒ P B A = 1 4 1 2 ⇒ P B A = 1 2 .

If A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 , then find P A B .

Given that A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 To find P A B : We know that P A B = P A ∩ B P B ⇒ P A B = 1 4 1 3 ⇒ P A B = 3 4 .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 112

EASY

12th ICSE

IMPORTANT

Earn 100

Can two events be mutually exclusive and independent simultaneously?

Important Questions on Probability

HARD

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 20

The probabilities that

A, B, C

can solve a problem independently are

\frac{1}{3}, \frac{1}{3}

and

\frac{1}{4}

respectively. Find the probability that only two of them are able to solve the problem

MEDIUM

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 22

A three digit number is written down by random choice of the digits

1

to

9

with replacements. The probability that atleast one of the digits chosen is a perfect square is

MEDIUM

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 25

A man draws a card from a pack of

52

cards and then replace it. After shuffling the pack, he again draws a card. This he repeats a number of times. The probability that he will draw a heart for the first time in the third draw is

MEDIUM

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 30

Let