P ( neither A nor B) =
= = =
=
There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits
Thus n(E) = 13 + 3 = 16
Hence
Therefore,
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 =
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
Let S be the sample space
Then, n(S) = = 1326.
Let E = event of getting 2 kings out of 4.
n(E) = = 6.
=>
There are 13 spade and 3 more jack
So probability of getting neither spade nor a jack: |
Let S be the sample space.
Here n(S)=
= 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.
Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.
Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events = 1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.
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