If A , B , C are three independent events of a random experiment such that P A ∩ B ¯ ∩ C ¯ = 1 4 , P A ¯ ∩ B ∩ C ¯ = 1 8 , P A ¯ ∩ B ¯ ∩ C ¯ = 1 4 , then find P A , P B and P C .

Given: A , B , C are independent events. P A ∩ B ¯ ∩ C ¯ = 1 4 ⇒ P ( A ) × P B ¯ × P C ¯ = 1 4 . . . . . . . . i P A ¯ ∩ B ∩ C ¯ = 1 8 ⇒ P A ¯ × P ( B ) × P C ¯ = 1 8 . . . . . . . . . . . . . . . ii P A ¯ ∩ B ¯ ∩ C ¯ = 1 4 ⇒ P A ¯ × P B ¯ × P C ¯ = 1 4 . . . . . . iii Divide equation i by equation iii , i iii = P ( A ) × P B ¯ × P C ¯ P A ¯ × P B ¯ × P C ¯ = 1 4 1 4 ⇒ P ( A ) P A ¯ = 1 ⇒ P ( A ) 1 - P ( A ) = 1 ⇒ P ( A ) = 1 - P ( A ) ⇒ 2 P ( A ) = 1 ⇒ P ( A ) = 1 2 Now, equation ( ii ) equation ( iii ) = P ( B ) P B ¯ = 1 8 1 4 = 1 2 ⇒ P ( B ) 1 - P ( B ) = 1 2 ⇒ 2 P ( B ) = 1 - P ( B ) ⇒ 3 P ( B ) = 1 ⇒ P ( B ) = 1 3 From equation i , P ( A ) × P B ¯ × P C ¯ = 1 4 ⇒ 1 2 1 - 1 3 P C ¯ = 1 4 ⇒ P C ¯ = 1 4 × 2 × 3 2 = 3 4 P ( C ) = 1 - P C ¯ = 1 - 3 4 = 1 4 ∴ P ( A ) = 1 2 , P ( B ) = 1 3 , P ( C ) = 1 4

Let X be a random variable such that P ( X = – 2 ) = P ( X = – 1 ) = P ( X = 2 ) = P ( X = 1 ) = 1 6 and P X = 0 = 1 3 . Find the mean and variance of X .

Given that: P ( X = – 2 ) = P ( X = – 1 ) = P ( X = 2 ) = P ( X = 1 ) = 1 6 and P ( X = 0 ) = 1 3 . Mean = ∑ x i P X = x i = - 2 × 1 6 + - 1 × 1 6 + 2 × 1 6 + 1 × 1 6 + 0 × 1 3 = - 2 - 1 + 2 + 1 × 1 6 + 0 Mean = 0 Now, variance = ∑ x i 2 P X = x i - mean 2 = - 2 2 × 1 6 + - 1 2 × 1 6 + 2 2 × 1 6 + 1 2 × 1 6 + 0 2 × 1 3 - 0 2 = 4 × 1 6 + 1 × 1 6 + 4 × 1 6 + 1 × 1 6 + 0 = 4 + 1 + 4 + 1 × 1 6 = 10 6 Variance = 5 3

If the difference between the mean and variance of a binomial variate is 5 9 , then find the probability for the event of 2 successes when the experiment is conducted five times.

Mean of the binomial distribution is n p and variance of binomial distribution is n p q , where p is the probability of success on any trial, q = 1 - p is the probability of failure, n is number of trials. Given: n = 5 and Mean - Variance = 5 9 ⇒ n p - n p q = 5 9 ⇒ n p ( 1 - q ) = 5 9 ⇒ n p 2 = 5 9 ⇒ 5 p 2 = 5 9 ⇒ p 2 = 1 9 ⇒ p = 1 3 q = 1 - p = 1 - 1 3 = 2 3 We know that, the formula for binomial distribution is P X = r = C r n p r q n - r , where p is the probability of success on any trail, q is the probability of failure, n is number of trials, r is the number of successes, it can take the values 0 , 1 , 2 , 3 , … n . Here, number of successes is r = 2 Now, P ( X = 2 ) = C 2 5 1 3 2 2 3 3 = 5 ! 2 ! 3 ! × 1 9 × 8 27 = 5 × 4 2 × 1 9 × 8 27 = 80 243 ∴ Probability for the event of 2 success = 80 243

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 33

HARD

12th Telangana Board

IMPORTANT

Earn 100

If the letters of the word $PRISON$ are permuted in all possible ways and the words thus formed are arranged in dictionary order, then find the rank of the word $PRISON$ .

Important Questions on Solved Mathematics Paper-1

HARD

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise-choice>Q 34

Find the numerically greatest terms in the expansion of ${(3 x - 4 y)}^{14}$ when $x = 8, y = 3$ .

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise-choice>Q 35

In a box containing $15$ bulbs, $5$ are defective. If $5$ bulbs are selected at random from the box, then find the probability of the event that none of them is defective.

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise-choice>Q 36

In a box containing $15$ bulbs, $5$ are defective. If $5$ bulbs are selected at random from the box, then find the probability of the event that only one of them is defective.

HARD

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise-choice>Q 37

In a box containing $15$ bulbs, $5$ are defective. If $5$ bulbs are selected at random from the box, then find the probability of the event that atleast one of them is defective

HARD

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise-choice>Q 38

If