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
Total number of cases=p+q
Favourable cases=p
Probability of that event is=p/(p+q)
Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
P (drawing a white ball) = 8 /14=4/7.
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) =
= =
Probability of atleast one failure
= 1 - no failure > 31/32
= 1 -
> 31/32
=
1/32
= p < 1/2
Also p > 0
Hence p lies in [0,1/2].
As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address
P(E) = favorable outcomes /total outcomes
Here favorable outcomes are '0'.
So probability is '0'.
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
P(E) =n(E)/n(S)=2/52=1/26.
Total numbers in a die=6
P(mutliple of 3) = 2/6 = 1/3
P(multiple of 4) = 1/6
P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2
Total number of events=52
Number of aces =4
So,required probability = 4/52 =1/13
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.