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:
= + + + = =
Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2
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 :
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 =
We know that:
When one card is drawn from a pack of 52 cards
The numbers of possible outcomes n(s) = 52
We know that there are 26 red cards in the pack of 52 cards
? The numbers of favorable outcomes n(E) = 26
Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)
? required probability = 26/52 = 1/2.
If 1 is adjacent to 2, 4 and 6 then either 3 or 5 lies opposite to 1. So, the numbers 3 and 5 cannot lie opposite to each other. Hence, 3 is adjacent to 5 (necessarily).
In a usual dice, the sum of the numbers on any two opposite faces is always 7. Thus, 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4.
Consequently, when 4, 3, 1 and 5 are the numbers on the top faces, then 3, 4, 6 and 2 respectively are the numbers on the face touching the ground. The total of these numbers = 3 + 4 + 6 + 2 = 15.
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