Here,n = 4(children)
P(girl)= 0.5
P(of atleast one girl)= 1 - P(no girls)
= 1 - 0.0625 = 0.9375
Number of queen cards = 4
Number of ace cards = 4
P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13
Total number of cases=36
Number of favourable cases(sum 5 or 6)=10
P(getting total of 5 or 6)=10/36=5/18
S = { 1, 2, 3, 4, 5, 6 }
=> n(S) = 6
E = { 3, 6}
=> n(E) = 2
Therefore, P(E) = 2/6 =1/3
The two events mentioned are independent.
The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.
P(getting first 4) = 1/6
P(no second 6) = 5/6
Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36
These two events cannot be disjoint because P(K) + P(L) > 1.
P(A?B) = P(A) + P(B) - P(A?B).
An event is disjoint if P(A ? B) = 0. If K and L are disjoint P(K ? L) = 0.8 + 0.6 = 1.4
And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.
4 persons can be selected from 9 in ways =126
Fvaourable events = =60
So,required probability = 60/126 = 10/21
Total number of elementary events = 52
There are 26 red cards,out of which one red card can be drawn in ways =26.
So,required probability = 26/52 = 1/2
Total number of elementary events = ways =190
There are 7 white balls out of which one white can be drawn in 7 ways and one ball from remaining 13 balls can be drawn in ways.
So,favourable events =
Therfore,required probability = ( )/ =91/90
Probability of getting 6 at the top once =1/6
Probability of getting 6 at the top three times =1/6 x 1/6 x 1/6 =216
Probability of no getting 6 at he top any time = 1 - 1/216 = 215/216
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= = 2300.
n(E)= = 1050.
P(E) = n(E)/n(s) = 1050/2300 = 21/46
Comments
There are no comments.Copyright İCuriousTab. All rights reserved.