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
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
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
4 persons can be selected from 9 in ways =126
Fvaourable events = =60
So,required probability = 60/126 = 10/21
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
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
Total numbers in die=6
P(multiple of 2)=1/2
P(multiple of 5)=1/6
P(multiple of 2 or 5)=2/3
P(heart cards)=13/52
P(diamond cards)=13/52
P(heart or diamond)=13/52+13/52=1/2
P(red cards)=26/52
P(black cards)=26/52
P(red or black cards)=26/52+26/52=1
P(red ball) = 5/12
P(white ball) = 4/12
P(red or white ball) = 5/12 + 4/12 = 3/4 = 0.75
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