Vowels are A I A I O,
C A S T I G A T I O N
(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)
So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :
Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2
P(odd) = P (even) = 1(because there are 50 odd and 50 even numbers)
Sum or the three numbers can be odd only under the following 4 scenarios:
Odd + Odd + Odd = =
Odd + Even + Even = =
Even + Odd + Even = =
Even + Even + Odd = =
Other combinations of odd and even will give even numbers.
Adding up the 4 scenarios above:
= + + + = =
Probability of occurrence of an event,
P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)
? Probability of getting head in one coin = ½,
? Probability of not getting head in one coin = 1- ½ = ½,
Hence,
All the 11 tosses are independent of each other.
? Required probability of getting only 2 times heads =
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.