The sample space of an experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH}, then A, B and C form a set of?
Above information has no common element in three of them and U of three sets forms sample set . Therefore option C is correct.
When the sets A and B are two events associated with a sample space, then "A ∩ B " is the event common to
Intersection of two sets A ∩ B is the set of those elements which are common to both A and B. i.e., which belong to both ‘A and B’.
If A and B are two events, then the set A ∩ B denotes the event ‘A and B’.
If an event has more than one sample point, it is called a ………
The numbers 2, 4, 6, 8 and 10 are written separately on five slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Then, the sample space for the experiment is:
Elements {2,4,6,8,10}
A person draws two slips from the box, one after the other, without replacement:
Sample Space ={(2,4) (2,6) (2,8) (2,10) (4,2) (4,6) (4,8) (4,10) (6,2) (6,4) (6,8) (6,10) (8,2) (8,4) (8,6) (8,10) (10,2) (10,4) (10,6) (10,8)}
If 2/7 is the probability of an event, then the probability of the event ‘not A’ is:
If 2/7 is probability of event A then the probability of an event not A is 1(2/7) i.e., probability of not A is 5/7
A possible result of a random experiment is called its ……
After the experiment, the result of the random experiment is known. An outcome is a result of a random experiment. The set of all possible outcomes is called the sample space.
Two students Amit and Priyanka appeared in an examination. The probability that Amit will qualify the examination is 0.06 and that Priyanka will qualify the examination is 0.12. The probability that both will qualify the examination is 0.02. Then, the probability that both Amit and Priyanka will not qualify the examination is:
Probability that Amit will qualify the exam(E) = 0.06
Probability that Priyanka will qualify the exam(F) = 0.12
Probability that both will qualify the exam P(E⋂F) = 0.02
P(EUF) = P(E) + P(F)  P(EUF)
= 0.06 + 0.12  0.02 = 0.16
P(E’⋂F’) = 1  P(EUF)
⇒ 1  0.16
= 0.84
The sample space associated to thrown a dice is
An experiment involves rolling a pair of dice and recording the number that comes up. Suppose,
A: the sum is greater than 8.
B: 2 occurs on either die. Then A and B are ……. events.
Mutually exclusive means no common event.
if 2 comes on either dive then sum can't be greater than 8.It will always less than or equal to 8.
So mutually exclusive.
A box contains 15 red marbles, 15 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that atleast one will be green?
Probability of getting atleast one green = 1 probability of getting no green = 1(30C_{5}/60C_{5}) = 4367/4484
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