n(S) = = 1326
Let A = event of getting both red cards
and B = event of getting both queens
then = event of getting two red queens
n(A) = = 325, n(B) = = 6
P ( both red or both queens) =
= =
S = { 1, 2, 3, 4, .....18 }
=> n(S) = 18
E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}
=> n(E1) = 9
E2 = {3, 6, 9, 12, 15, 18 }
=> n(E2) = 6
=> n(E3) = 3
=> n(E) = 9 + 6 - 3 =12
where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }
Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events = 1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.
Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.
Let S be the sample space.
Here n(S)=
= 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.
There are 13 spade and 3 more jack
So probability of getting neither spade nor a jack: |
Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
P (drawing a white ball) = 8 /14=4/7.
Total number of cases=p+q
Favourable cases=p
Probability of that event is=p/(p+q)
Let A be the event of driver A drive safely after consuming liquor.
Let B be the event of driver B drive safely after consuming liquor.
Let C be the event of driver C drive safely after consuming liquor.
P(A)=2/5,P(B)=3/7,P(C)=3/4
The events A, B and C are independent . Therefore,
Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70
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