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
Total numbers = 25
Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9
The probability (divisible by 4 or 7) = 9/25
S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }
=> n(S) = 8
E = { HHH, HHT, HTH, THH }
=> n(E) = 4
P(E) = 4/8 = 1/2
Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18
Multiples of 5 below 20 are 5, 10, 15, 20
Required number of possibilities = 10
Total number of possibilities = 20
Required probability = 10/20 = 1/2.
Let A be the event of A will tell truth. B be the event of B tell truth
When both agree then they say true or they say false together, that is
Also these events will be mutually exclusive :
Total number of events=52
Number of aces =4
So,required probability = 4/52 =1/13
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
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
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
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
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