Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then
Required probability =
=
Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9
When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white
Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.
Hence the required probability =
Required probability is given by P(E) =
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
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
Total no of ways = (14 ? 1)! = 13!
Number of favorable ways = (12 ? 1)! = 11!
So, required probability = = =
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
=
Let A be the event that X is selected and B is the event that Y is selected.
P(A) = 1/7, P(B) = 2/9.
Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
= (1/7) × (2/9) = 2/63
Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)
=
= = =
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
The possible outcomes are as follows :
5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.
Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).
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