We have n(s) = 52 = 52*51/2*1= 1326.
Let A = event of getting both black cards
B = event of getting both queens
A?B = event of getting queen of black cards
n(A) = = = 325, n(B)= = 4*3/2*1= 6 and n(A?B) = = 1
P(A) = n(A)/n(S) = 325/1326;
P(B) = n(B)/n(S) = 6/1326 and
P(A?B) = n(A?B)/n(S) = 1/1326
P(A?B) = P(A) + P(B) - P(A?B) = (325+6-1) / 1326 = 330/1326 = 55/221
Let A, B, C be the respective events of solving the problem and be the respective events of not solving the problem. Then A, B, C are independent event
are independent events
Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4
P( none solves the problem) = P(not A) and (not B) and (not C)
=
=
=
=
Hence, P(the problem will be solved) = 1 - P(none solves the problem)
= = 3/4
Let S be the sample space
Then n(S) = no of ways of drawing 2 balls out of (6+4) = 10 = =45
Let E = event of getting both balls of same colour
Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)
= = = 15+6 = 21
Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20.
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36 = 5/12.
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364days
366 ? 364 = 2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days = 2/7
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