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S seema garhwal
X be number od aces obtained. X can be 0,1,2 There 52 cards and 4 aces, 48 are non-ace cards. The probability distribution is as : X 0 1 2 P(X)   Option D is correct.

S seema garhwal
X is number representing on die.  Total observations = 6 X 1 2 5 P(X) Option B is correct.

S seema garhwal
Given :                       Probability distribution is as : X 0 1 P(X) 0.3 0.7

S seema garhwal
Total students = 15 probability of selecting a student :                                                            The information given can be represented as frequency table : X 14 15 16 17 18 19 20 21 f 2 1 2 3 1 2 3 1                                                                                                                                                                    ...

S seema garhwal
denote the sum of the numbers obtained when two fair dice are rolled. Total observations = 36 X can be 2,3,4,5,6,7,8,9,10,11,12 Probability distribution is as follows :  X 2 3 4 5 6 7 8 9 10 11 12 P(X)                                                                          Standard deviation =

S seema garhwal
Two numbers are selected at random (without replacement) from the first six positive integers in  ways. denote the larger of the two numbers obtained. X can be 2,3,4,5,6. X=2, obsevations :  X=3, obsevations :  X=4, obsevations :  X=5, obsevations :  X=6, obsevations :  Probability distribution is as follows:         X 2 3 4 5 6 P(X)

S seema garhwal
denotes the number of sixes, when two dice are thrown simultaneously. X can be 0,1,2.     Not getting six on dice         Getting six on one time when  thrown twice :          Getting six on both dice :  X 0 1 2 P(X) Expectation of X = E(X)

S seema garhwal
Let X be success of getting head. When 3  coins are  tossed  then sample space    X can be 0,1,2,3 The probability distribution is as  X 0 1 2 3 P(X) mean number of heads :

S seema garhwal

S seema garhwal
Sum of probabilities of probability distribution of random variable is 1.

S seema garhwal

S seema garhwal

S seema garhwal

S seema garhwal
Sum of probabilities of probability distribution of random variable is 1.

S seema garhwal
the coin is tossed twice, total outcomes =4  probability of getting a tail be x. i.e.  Then                                                                                          and            Let  X : number of tails No tail :  1 tail :  2 tail :  the probability distribution of number of tails are  X 0 1 2 P(X)

S seema garhwal
Total bulbs = 30 defective bulbs = 6 Non defective bulbs  bulbs is drawn at random with replacement. Let X : number of defective bulbs  4 Non defective bulbs and 0  defective bulbs :  3 Non defective bulbs and 1  defective bulbs :  2 Non defective bulbs and 2  defective bulbs :  1 Non defective bulbs and 3  defective bulbs :  0 Non defective bulbs and 4  defective bulbs :  the probability...

S seema garhwal
When a die is tossed twice , total outcomes = 36 Six does not appear on any of the die  :   Six appear on atleast one die  :  Probability distribution is as :         X 0 1 P(X)

S seema garhwal
When a die is tossed twice , total outcomes = 36 Number less than or equal to 4 in both toss  :   Number less than or equal to 4 in first toss and number more than or equal to 4 in second toss + Number less than or equal to 4 in second toss and number more than or equal to 4 in first toss:  Number less than 4 in both tosses :  Probability distribution is as : X 0 1 2 P(X)