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
Total balls = 40
Red balls = 18
Let green balls are x
Then, (18/40) × (
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 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.
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.
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
=
Total no of ways = (14 ? 1)! = 13!
Number of favorable ways = (12 ? 1)! = 11!
So, required probability = = =
A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}
Required Probability = 1/7
As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).
The overall resultant will remain same.
So final amount with the person will be (in all cases):
= 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27
Hence the final result is:
64 ? 27 = 37
A loss of Rs.37
Required probability is given by P(E) =
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