Total number =6
Getting a 'multiple of 3' = 2 . So, probability = 2/6 =1/3
The number of exhaustive events = 100 C? = 100.
We have 25 primes from 1 to 100.
Number of favourable cases are 75.
Required probability = 75/50 = 3/2.
S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)} xy will be even when even x or y or both will be even. P(E)=n(E)n(S)=812 = 4/3 |
We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and
E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)
n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36
Total number of possible ways = 9C3 = 84 ways
Required atleast one girl in the group of three = total possible ways - ways in which none is girl
None of the members in the group is girl = 6C3 = 20
Therefore, number of ways that at least one member is a girl in the group of three
= 84 - 20
= 64 ways.
Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.
So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}
Favourable number of cases=10
Therefore,P(B)=10/216
=> P(A) = 1 - P(B)
= 1 - (10/216) = 103/108
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.
Here in this pack of cards, n(S) = 52
Let E = event of getting a queen of the club or a king of the heart
Then, n(E) = 2
P(E) = n(E)/n(S) = 2/52 = 1/26
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