In two throws of a die, n(s)=(6 x 6)=36
let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}
P(E) = n(E)/n(S) = 4/36 = 1/9
250 numbers between 101 and 350 i.e. n(S)=250
n(E)=100th digits of 2 = 299?199 = 100
P(E)= n(E)/n(S) = 100/250 = 0.40
Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16
Required probability =(12)×(16)
= 1/12
Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4
Total number of persons = 9
Out of 9 persons 4 persons can be selected in ways =126
1 man,1 woman and 2 children can be selected in
ways =36
So,required probability = 36/126 =2/7
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.
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
S = { 1, 2, 3, 4, 5, 6 }
=> n(S) = 6
E = { 3, 6}
=> n(E) = 2
Therefore, P(E) = 2/6 =1/3
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
Number of queen cards = 4
Number of ace cards = 4
P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13
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