X can take 7 values.
To get |X|+2) take X={?1,0,1}
=> P(|X|<2) = Favourable CasesTotal Cases = 3/7
Here n(S) = (6 x 6) = 36
Let E = event of getting a total more than 7
= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.
Here n(S) = 6 x 6 = 36
E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}
=> n(E)=20
Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.
Probability is Number of favourable outcomes by total outcomes
Total comes = 2 (Divisible or not)
Favourable = 1 (Divisible)
Hence, probability = 1/2.
Probability between z = 0 and z = 3.01 is given by
P(0<z<3.01) = P(z<3.01) - P(z<0)
Reading from the z-table, we have
P(z<0) = 0.5
P(z<3.01) = 0.9987
Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.
The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
=
n(S) =
Let X be the event that all dice show the same face.
X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}
n(X) = 6
Hence required probability = =
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7 = = 21
Let E = Event of drawing 2 balls, none of which is blue.
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = = 10
Therefore, P(E) = n(E)/n(S) = 10/ 21.
Here S = {1,2,3,4,5,6}
Let E be the event of getting the multiple of 3
Then, E = {3,6}
P(E) = n(E)/n(S) = 2/6 = 1/3
Total balls = 40
Red balls = 18
Let green balls are x
Then, (18/40) × (
Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.
Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.
Required probability =24/64 = 3/8
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
=
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