If A and B are two events such that P ( A ∪ B ) = 0 . 7 , P ( A ∩ B ) = 0 . 2 , and P ( B ) = 0 . 5 , then show that A and B are independent.

Given A and B are two events such that P ( A ∪ B ) = 0 . 7 , P ( A ∩ B ) = 0 . 2 and P ( B ) = 0 . 5 We know that, for two independent events A and B , P ( A ∩ B ) = P ( A ) · P ( B ) and P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) Now, 0 . 7 = P ( A ) + 0 . 5 - 0 . 2 ⇒ P ( A ) = 0 . 7 - 0 . 3 = 0 . 4 ⇒ P ( A ) · P ( B ) = 0 . 4 × 0 . 5 = 0 . 20 ∴ P ( A ∩ B ) = P ( A ) . P ( B ) = 0 . 20 ∴ A and B are independent.

If A and B are two independent events such that P ( A ∪ B ) = 0 . 6 , P ( A ) = 0 . 2 find P ( B ) .

Given that, A and B are two independent events such that P ( A ∪ B ) = 0 . 6 , P ( A ) = 0 . 2 Since, A and B are independent, we have 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 . 6 = 0 . 2 + P ( B ) - 0 . 2 P ( B ) ⇒ 0 . 6 - 0 . 2 = ( 1 - 0 . 2 ) P ( B ) ⇒ P ( B ) = 0 . 4 0 . 8 ∴ P ( B ) = 1 2 = 0 . 5 .

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

If P A = 7 / 13 , P B = 9 / 13 and P A ∩ B = 4 / 13 then find P ( A | B ) .

If P ( B ) = 0 . 5 and P ( A ∩ B ) = 0 . 32 , then find P ( A | B ) .

Given: P ( B ) = 0 . 5 and P ( A ∩ B ) = 0 . 32 To find: P A | B We know that P A | B = P A ∩ B P B ⇒ P ( A | B ) = 0 . 32 0 . 5 ⇒ P ( A | B ) = 32 50 ⇒ P ( A | B ) = 16 25 Hence, the probability P A | B is 16 25 .

If P ( A ) = 0 . 6 , P ( B ) = 0 . 3 and P A ∩ B = 0 . 2 , then find P A | B and P B | A .

If P A = 0 . 8 , P ( B ) = 0 . 5 and P B | A = 0 . 4 , then find P ( A ∩ B ) .

Given: P A = 0 . 8 , P B = 0 . 5 and P B | A = 0 . 4 To find: P A ∩ B We know that P B | A = P A ∩ B P A ⇒ P A ∩ B = P B | A × P A ⇒ P ( A ∩ B ) = 0 . 4 × 0 . 8 ⇒ P ( A ∩ B ) = 0 . 32 Hence, the probability P ( A ∩ B ) is 0 . 32 .

Ten cards numbered 1 to 10 are placed in box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3 , what is the probability that it is an even number ?

Ten cards numbered 1 to 10 are placed in a box and one card is drawn randomly. To find: Probability that number on card drawn is even given that the number is more than 3 Let A : event that number on card drawn is even number and B : event that number on card drawn is more than 3 . Sample space S = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 . Then, A = 2 , 4 , 6 , 8 , 10 B = 4 , 5 , 6 , 7 , 8 , 9 , 10 So, A ∩ B = 4 , 6 , 8 , 10 Thus, we get P A = 5 10 , P B = 7 10 and P A ∩ B = 4 10 We know that P A | B = P A ∩ B P B ⇒ P ( A | B ) = 4 / 10 7 / 10 = 4 7 which is the required probability.

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 .