The probability for a contractor to get a road contract is 2 3 and to get a building contract is 5 9 . The probability to get at least one contract is 4 5 . Find the probability that he gets both the contracts.

Let A is the event of getting a road contract. B is the event of getting a building contract. ∴ P ( A ) = 2 3 , P ( B ) = 5 9 , P ( A ∪ B ) = 4 5 We know that, from addition theorem of probability P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) ⇒ 4 5 = 2 3 + 5 9 - P ( A ∩ B ) ⇒ P A ∩ B = 2 3 + 5 9 - 4 5 = 30 + 25 - 36 45 = 19 45 Therefore, the probability to get both contract = 19 45 .

State and prove Bayes' theorem.

The statement of Bayes' theorem is as follows: Let E 1 , E 2 , E 3 , . . . . . , E n be a set of events associated with a sample space S , where all events E 1 , E 2 , E 3 , . . . . . , E n have a non-zero probability of occurrence and they form a partition of S . Let A be any event which occurs with E 1 or E 2 or E 3 . . . . . or E n , then according to Bayes Theorem, P E i / A = P E i P A / E i ∑ k = 1 n P E k P A / E k , i = 1 , 2 , 3 , . . . . , n Here E i ∩ E j = ϕ , where i ≠ j , i.e. they are mutually exhaustive events. The union of all the events of the partition should give the sample space. To prove the Bayes' theorem, we will use the total probability and conditional probability formulas. E i be is a partition of the sample space S . Let A be an event that occurred. According to conditional probability, P E i / A = P E i ∩ A P A . . . . . . . 1 Using the multiplication rule of probability, P E i ∩ A = P E i P A / E i Here, P A = P E 1 ∩ A + P E 2 ∩ A + . . . . + P E n ∩ A ⇒ P A = P E 1 P A / E 1 + P E 2 P A / E 2 + . . . . + P E n P A / E n From the total probability theorem, ⇒ P A = ∑ k = 1 n P E k P A / E k Substituting values in equation 1 , P E i / A = P E i P A / E i ∑ k = 1 n P E k P A / E k , i = 1 , 2 , 3 , . . . , n . Hence, proved.

The probability for a contractor to get a road contract is 2 3 and to get a building contract is 5 9 . The probability to get at least one contract is 4 5 . Find the probability that he gets both the contracts.

Let A is the event of getting a road contract. B is the event of getting a building contract. ∴ P ( A ) = 2 3 , P ( B ) = 5 9 , P ( A ∪ B ) = 4 5 We know that, from addition theorem of probability P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) ⇒ 4 5 = 2 3 + 5 9 - P ( A ∩ B ) ⇒ P A ∩ B = 2 3 + 5 9 - 4 5 = 30 + 25 - 36 45 = 19 45 Therefore, the probability to get both contract = 19 45 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Probability>Andhra Pradesh Board-2018>Q 4

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

Earn 100

State and prove Addition theorem on probability.

Important Questions on Probability

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Probability>Andhra Pradesh Board-2018>Q 5

Find the probability of drawing an ace or a spade from a well shuffled pack of

52

.

EASY

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Probability>Andhra Pradesh Board-2019>Q 6

The probabilities of three mutually exclusive events are respectively given as

\frac{1 + 3 p}{3}, \frac{1 - p}{4}, \frac{1 - 2 p}{2} .

Prove that

\frac{1}{3} ⩽ p ⩽ \frac{1}{2} .

MEDIUM

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Probability>Andhra Pradesh Board-2019>Q 7

If

A

and

B

are independent events of a random experiment, show that

A^{C}

and

B^{C}

are also independent.

HARD

12th Andhra Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 9 - Probability>Andhra Pradesh Board-2019>Q 8

Suppose that an urn $B_{1}$ contains $2$ white and $3$ black balls and another urn $B_{2}$ contains $3$ white and $4$ black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the probability that the urn chosen was $B_{1}$ .