Question 1 :
If two coins are tossed simultaneously, then find the probability of getting (i) one head and one tail (ii) at most two tails
Solution :
Sample space = { HH, HT, TH, TT }
n(S) = 4
(i) one head and one tail
Let "A" be the event of getting the probability of one head and one tail.
A = {HT, TH}
n(A) = 2
P(A) = n(A) / n(S)
= 2/4 ==> 1/2
(ii) at most two tails
Let "B" be the event of getting the probability of at most two tails.
At most two tails means, we may get 0 tail, 1 tail and 2 tails.
B = { HH, HT, TH, TT }
n(B) = 4
P(B) = n(B) / n(S)
= 4/4 ==> 1
Question 2 :
Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that (i) one is a mango and the other is an apple (ii) both are of the same variety.
Solution :
Total number of fruits in a box = 5 mangoes + 4 apples
= 9
Since we take two fruits from the box, the sample space will be ^{9}C_{2}
^{9}C_{2 } = (9 ⋅ 8)/(2 ⋅ 1) = 36
(i) one is a mango and the other is an apple
Let "A" be the event of getting one is a mango and other is an apple.
Thus we select 1 mango from 5 mangoes and 1 apple from 4 apples.
n(A) = ^{5}C_{1 }⋅ ^{4}C_{1}
n(A) = 5 ⋅ 4 = 20
P(A) = n(A)/n(S)
= 20/36
P(A) = 5/9
(ii) both are of the same variety.
Let "B" be the event of getting the probability of both are of the same variety.
n(B) = (^{5}C_{2 }⋅ ^{4}C_{0}) + (^{5}C_{0 }⋅ ^{4}C_{2})
= 10 + 6
n(B) = 16
P(B) = n(B)/n(S)
= 16/36
p(B) = 4/9
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