Test: Probability And Statistics - 1


20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Test: Probability And Statistics - 1


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QUESTION: 1

Consider the experiment of throwing two fair dice. What is the probability that the sum of the numbers obtained in these dice is even?

Solution:

As sum of two odd numbers and sum of two even numbers is even, so required probability is,

QUESTION: 2

Two dice are rolled until the sum of the numbers appearing on these dice is either 7 or 8. What is the probability that the sum is 7?

Solution:

Sum of the numbers on the faces of t wo dice is 7 in 6 cases and 8 in 5 cases., so by conditional probability’s theorem of Bayes the sum obtained will be 7 has probability,

QUESTION: 3

A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is.

Solution:

Number of cases for sum ‘5’ = 4 
Number of cases for sum ‘7’ = 6 
Probability of getting sum 5 in one roll = 4/36 = 1/9
Probability' of getting neither 5 nor 7 in a roll

⇒ Probability of getting 5 before 7

QUESTION: 4

A letter is taken at random from the letters of the word ‘STATISTICS ’ and an other letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that, they are the same letter is .

Solution:

The same letter can be either, S,T, I or A.
Probability required will be.

QUESTION: 5

A bag contains 6 red and 4 green balls. A fair dice is rolled and a number of balls equal to that appearing on the dice is chosen from the urn at random. The probability that all the balls selected are red is.

Solution:

Required probability =

QUESTION: 6

A six faced die is a biased one. it is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even, is.

Solution:


Now sum of two odd numbers, and also sum of two even numbers is even, so.
P (odd, odd) + P (even, even)

QUESTION: 7

A letter is knwon to have come from either TATANAGAR' or CALCUTTA. On the envelope, Just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is.

Solution:

TATANAGAR has 9 letters, so number of ways in which two consecutive letters can be printed will be, 8 out of which there are 2 ways in which TA can be printed.
Similarly for CALCUTTA, there are 7 ways of printing two consecutive letters, from which there is only way to print TA'
So, required probability’

QUESTION: 8

A card is drawn from a pack. The card is replaced and the pack is reshuffled. If this is done six times, the probability that 2 hearts, 2 diamonds and 2 club cards are drawn is.

Solution:

Probability of selecting card of any suit = 
so, due to replacement probability that out of 6 cards drawn there will be 2 Hearts, 2 diamonds and 2 club cards is

QUESTION: 9

There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls is.

Solution:

Possible configuration of balls to be selected seven possible cases are there.

QUESTION: 10

Two balls are drawn at random from a bag containing 6 white, 4 red and 5 black balls. The probability that both these balls are black, is :

Solution:

 Required probability = 

QUESTION: 11

Let X be the random variable giving the number of heads obtained in 162 successive tosses of a biased coin with probability of getting head in a toss is 1/3.  Assume that the losses are in dependent. The standard deviation of X is.

Solution:

Random variable x, follows Binomial distribution with n = 162, p = 1/3so, q = 1 - p = 2/3
variance = npq = 162(1/3)(2/3) = 36
⇒ standard deviation = √variance = 6

QUESTION: 12

Assume that the number of buses arriving at a bus-stop in an interval of t minutes is poisson distributed with parameter 0.21. The probability that at least three buses will arrive during a 10 minute interval is.

Solution:

Probability mass function of poission distribution with parameter 0.2t is

QUESTION: 13

The standard deviation of first, n natural numbers is :

Solution:

QUESTION: 14

Of a distribution, coefficient of skewness is 0.32, standard deviation is 6.5 and the mean is 29.6 then its mode will be :

Solution:

QUESTION: 15

A ship arrives at a port and 40 sailors on board go ashore for revelry. Later at night, the 40 sailors return to the ship and in their state of inebriation, each chooses a random cabin to sleep in. What is the expected number of sailors sleeping in their own cabin?

Solution:

probability that the sailor returns in its own cabin is 1/40 If sailor is in his own cabin in  the expected value head count will be increased by 1.

QUESTION: 16

Suppose there are 15 different types of coupons and suppose that each time one obtains a coupon, it is equally likely to be any one of the 15 types. The expected number of different types that are contained in the set of 5 coupons is

Solution:

Probability that any particular coupon will be selected is 1/15.
Expected number of different types of coupons = 

QUESTION: 17

The joint density of X and Y is given by The value of E|X/2|Y = 1 is equal to

Solution:


Marginal density function of y

QUESTION: 18

A and B throw with one die for a stake of Rs 11 which is to be won by the player who first throws 6. If A has the first throw, what is the expectation of his winning the stake?

Solution:

Probability of obtaining 6 in a single throw is 1/6. Probability that A wins the stake if he get 6 in first throw or if he loses then B also lose and then he get 6 in his second throw i.e. overal 3rd throw and so on.

 [sum of infinite terms of G.P.]

QUESTION: 19

If the random variables X, Y and Z have the means μx = 2, μy= -3 and μz = 2, the variances  and covariances cov(X,Y) = -2, cov(X, Z) = -1 and cov (Y, Z) = 1, the variance of W = 3X - Y+ 2Z is

Solution:

W = 3 X - Y + 2 Z

QUESTION: 20

A computationally efficient way to compute the sample mean of the data x1,x2,...xn is as follows:

Then K(j) is equal to

Solution:


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