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A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \(P(A \cup B)\) is

Solution : A = {4. 5. 6}  and  B = {1, 2, 3, 4} \(A \cap B\) = 4 \(\therefore\)  \(P(A \cup B)\) = P(A) + P(B) – \(A \cap B\) \(\implies\) \(P(A \cup B)\) = 3/6 + 4/6 – 1/6 = 1 Similar Questions The probability of India winning a test match against the …

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \(P(A \cup B)\) is Read More »

It is given that the events A and B are such that P(A) = 1/4, P(A/B) = 1/2 and P(B/A) = 2/3. Then, P(B) is equal to

Solution : We know that, P(A/B) = \(P(A \cap B)\over P(B)\)    …….(i) and P(B/A) = \(P(B \cap A)\over P(A)\)    ……….(ii) \(\therefore\)  P(B) = \(P(B/A).P(A)\over P(A/B)\) = \(1\over 3\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that …

It is given that the events A and B are such that P(A) = 1/4, P(A/B) = 1/2 and P(B/A) = 2/3. Then, P(B) is equal to Read More »

Find the variance of first n even natural numbers ?

Solution : \(\therefore\) Variance = [\({1\over n}\sum{(x_i)^2}\)] – \((\bar{x})^2\) = \(1\over n\)[\(2^2 + 4^2 + ….. + (2n)^2\)] – \((n+1)^2\) = \(1\over n\)\(2^2 [ 1^2 + 2^2 + ….. + n^2]\) – \((n+1)^2\) = \(4\over n\) \(n(n + 1)(2n + 1)\over 6\) – \((n+1)^2\) =  \((n + 1)[(2n + 1) – 3(n + 1)\over 3\) …

Find the variance of first n even natural numbers ? Read More »

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ……, 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero equal to

Solution : S = { 00, 01, 02, ……, 49 } Let A be the event that sum of the digits on the selected ticket is 8, then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero. B = { 00, 01, …

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ……, 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero equal to Read More »

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is

Solution : Total number of cases = \(^9C_3\) = 84 Number of favourable cases = \(^3C_1\).\(^4C_1\).\(^2C_1\) = 24 \(\therefore\)  P = \(24\over 84\) = \(2\over 7\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series …

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is Read More »

A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 g and a standard deviation of 2 g. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 g. The correct mean and standard deviation in gram of fishes are respectively.

Solution : Correct mean = old mean + 2 = 30 + 2 = 32 As standard deviation is independent of change of origin. Hence, it remains same. Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is The median …

A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 g and a standard deviation of 2 g. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 g. The correct mean and standard deviation in gram of fishes are respectively. Read More »

Let A, B and C are pairwise independent events with P(C) > 0 and \(P(A \cap B\cap C)\) = 0. Then \(P(A’ \cap B’/C)\) is equal to

Solution : \(P({A’ \cap B’\over C})\) = \(P(A’ \cap B’ \cap C)\over P(C)\) = \(P(C) – P(A \cap C) – P(B \cap C) + P(A \cap B\cap C)\over P(C)\)   ……..(i) Given, \(P(A \cap B\cap C)\)  = 0 and A, B and C are pairwise independent. \(\therefore\)  \(P(A \cap C)\) = P(A).P(C) and \(P(B \cap C)\) …

Let A, B and C are pairwise independent events with P(C) > 0 and \(P(A \cap B\cap C)\) = 0. Then \(P(A’ \cap B’/C)\) is equal to Read More »

If the mean deviation about the median of numbers a, 2a, …. , 50a is 50, then |a| is equal to

Solution : Median of a, 2a, 3a, 4a, ….. . 50a is \(25a + 26a\over 2\) = 25.5a Mean deviation = \(\sum{|x_i – Median|}\over N\) \(\implies\)  50 = \(1\over 50\) {2|a|.(0.5 + 1.5 + …… + 24.5)] \(\implies\) 2500 = 2|a|. \(25\over 2\) (25) \(\implies\) |a| = 4 Similar Questions The mean and variance of …

If the mean deviation about the median of numbers a, 2a, …. , 50a is 50, then |a| is equal to Read More »

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

Solution : Probability of guessing a correct answer, p = \(1\over 3\) and probability of guessing a wrong answer, q  = \(2\over 3\) So, the probability of guessing 4 or more correct answers is = \(^5C_4\) \(({1\over 3})^4\). \(2\over 3\) + \(^5C_5\) \(({1\over 3})^5\) = \(5.2\over {3^5}\) + \(1\over {3^5}\) = \(11\over {3^5}\) Similar Questions …

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is Read More »

All the students of a class performed poorly in mathematics. The teacher decided to give grace marks of 10 to each of the students. Which statistical measure will not change even after the grace marks were given ?

Question : All the students of a class performed poorly in mathematics. The teacher decided to give grace marks of 10 to each of the students. Which statistical measure will not change even after the grace marks were given ? (a) Mean (b) Median (c) Mode (d) Variance Solution : If initially all marks were …

All the students of a class performed poorly in mathematics. The teacher decided to give grace marks of 10 to each of the students. Which statistical measure will not change even after the grace marks were given ? Read More »

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