Maths Questions

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is

Solution : Probability of getting success, p = \(1\over 6\) and probability of failure, q = \(5\over 6\) Now, we must get 2 sixes in seven throws, so probability is \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\) and the probability that 8th throw is \(1\over 6\). \(\therefore\)   Required Probability = \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)\(1\over 6\) = \(^7C_2\times 5^5\over {6^8}\) Similar โ€ฆ

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is Read More ยป

If A and B are two mutually exclusive events, then

Question : If A and B are two mutually exclusive events, then (a) P(A) < P(Bโ€™) (b) P(A) > P(Bโ€™) (c) P(A) < P(B) (d) None of these Solution : Since, A and B are two mutually exclusive events. \(\therefore\)    A \(\cap\) B = \(\phi\) \(\implies\) Either A \(\subseteq\) Bโ€™ or B \(\subseteq\) Aโ€™ โ€ฆ

If A and B are two mutually exclusive events, then Read More ยป

A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is

Solution : The total number of ways in which numbers can be choosed = 25*25 = 625 The number of ways in which either players can choose same numbers = 25 \(\therefore\) Probability that they win a prize = \(25\over 625\) = \(1\over 25\) Thus, the probability that they will not win a prize in โ€ฆ

A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is Read More ยป

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?

Solution : Since, total number of students = 100 and number of boys = 70 \(\therefore\) number of girls = (100 โ€“ 70) = 30 Now, the total marks of 100 students = 100*72 = 7200 And total marks of 70 boys = 70*75 = 5250 Total marks of 30 girls = 7250 โ€“ 5250 โ€ฆ

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? Read More ยป

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is \(1\over 2\), \(1\over 3\) and \(1\over 4\). Probability that the problem is solved is

Solution : Since the probability of solving the problem by A, B and C is \(1\over 2\), \(1\over 3\) and \(1\over 4\) respectively. \(\therefore\)  Probability that problem is not solved is = P(Aโ€™)P(Bโ€™)P(Cโ€™) = (\(1 โ€“ {1\over 2}\))(\(1 โ€“ {1\over 3}\))(\(1 โ€“ {1\over 4}\)) = \(1\over 2\)\(2\over 3\)\(3\over 4\) = \(1\over 4\) = Hence, the โ€ฆ

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is \(1\over 2\), \(1\over 3\) and \(1\over 4\). Probability that the problem is solved is Read More ยป

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

Solution : Given that, for binomial distribution mean, np = 4 and variance, npq = 2. \(\therefore\)  q = 1/2, but p + q = 1 \(\implies\) p = 1/2 and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8 We know that,  P(X = r) = \(^nC_r p^r q^{n-r}\) \(\therefore\)  P(X = 1) โ€ฆ

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is Read More ยป

Five horses are in a race. Mr A selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is

Solution : The probability that Mr A selected the lossing horse = \(4\over 5\) \(\times\) \(3\over 4\) = \(3\over 5\) The probability that Mr A selected the winning horse = 1 โ€“ \(3\over 5\) = \(2\over 5\) Similar Questions Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an โ€ฆ

Five horses are in a race. Mr A selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is Read More ยป

Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an even sum.

Solution : Sample space of rolling two dice = 36 Now, Event of sum of 5 = { (1,4) (2,3) (3,2) (4,1) } = 4 \(\implies\) probability of getting sum of 5 is 4/36 = 1/9 Now, Event of even sum = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) โ€ฆ

Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an even sum. Read More ยป

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is

Solution : Median of new set remains the same as of the original set. 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 Let \(x_1\), \(x_2\), โ€ฆ.. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) โ€ฆ

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is Read More ยป

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is

Solution : Given that, mean = 4 \(\implies\) np = 4 And Variance = 2 \(\implies\) npq = 2 \(\implies\) 4q = 2 \(\implies\)  q = \(1\over 2\) \(\therefore\)   p = 1 โ€“ q = 1 โ€“ \(1\over 2\) = \(1\over 2\) Also, n = 8 Probability of 2 successes = P(X = 2) = โ€ฆ

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is Read More ยป

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