Example 81
A die is thrown twice and sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 2 has appeared at least once ?
Sol. n (S) = 36. Let A denote the event 'a sum of T then A = (1,6), (2,5), (3,4), (4,3), (5,2), (6,-l) n(A) = 6
Let B denote the event 'the number 2 has appeared at least once'.
B = (1,2),(2,1), (2,2),(2,3), (2,4), (2,5), (2,6),(3,2), (4,2), (5,2), (6,2)
ADB = (2,5),(5,2) .-. n(AClB) = 2 => P(A HB)= " = ^= ^
A) n(S)~ 36~6 P(/A) P (A) 1/6 3'
Example 82
IfP(E) = 0.40,P(F) = 035 andP(E(JF) = 0.55, find P(EIF). '
Sol. P (E) = 0.40, P(F) = 0.35, P (Ef]F) = 0.55
P(E\jF) = P(E) + P(F)-P(EC)F) ' „
035 = OAO + 035-P(EnF)=>P(EnF) = 0.20
PCE/F)' = ^Wjlm.l ^ ; P(F) 0.35 7
Example 83
A bag contains 3 red and 4 black balls and another bag has 4 red and 2 black balls. One bag is selected, each of the two bags being equally to be selected. From the selected bag a ball is drawn, each ball in the bag being equally likely to be drawn. Let £ be the event "The first bag is selected, F be the event "The second bag is selected" and G the event "The ball is drawn red." Find P(E),P(F),P(GIE),P(GIF).
Sol. />(£)= I.p^-I,
P(GiE) = Probability .of drawing a red ball when first is selected
3
= Probability of drawing a red ball from first bag = ~
P (GIF) = Probability of drawing a red ball from second bag = 7 = 4 • • - u * 6 13 1
Example 84
Two integers are selected at random from integers 1 through 11. If the sum is even, find the probability that both the numbers are odd.
Sol. The integers from 1 through 11 are 1,2,3,4,5,6,7,8,9,10,11. Out of these, there are 5 even and 6 odd integers. ' '
Let A be the event 'both numbers chosen are odd' and B be the event' the sum of the numbers 6C 3
chosen is even. ThenP(v4)= ^-2- = — . Now, since the sum oftwo integers is even ifeither both the chosen integers are even or both are odd, therefore,
2
r, . . 0(Am. P(AnB) 3/11 3
k. Required probability = P(A/B) = ~= 5777 = 5 •
Example 85
Find P (B/A), if (i) A is a subset of B, (ii) A and B are mutually exclusive.
Sol.
(i) If A is a subset of B, then whenever A occurs B must occur, hence P (B/A) = 1
P (A 0 B) P (A)
Alternately, if A is a subset of B then A C\B = A\ hence P (B/A) = —TTTTv— ~ ~~pT~a\ = 1 •
r \A) r \A)
(ii) If A and B are mutually exclusive, i.e., disjoint, then whenever^ occurs B cannot occur, hence P (B/A) - 0. Alternately, if A and B are mutually exclusive then 1
Example 86
In a certain college, 25% of the students failed in Mathematics, 15% of the students failed in chemistry and 10% of the students failed both in Mathematics and chemistry. A student is selected at random
(i) If he failed in Chemistry, what is the probability that he failed in Mathematics ?
(ii) If he failed in Mathematics, what is the probability that he failed in Chemistry ?
(iii) What is the probability that he failed in Mathematics or Chemistry ?
Sol. Let A : Student failed in Mathematics ^ B: Student failed in Chemistry * ^ t
P(A)= ^ ^0.25,/>(*) = ^ =0.15, P(AC\B)= ^ =0.10 (0 " Required probability = = = |
(») Required probability = P (B/A) = —JT^— ~~ q^J ~ 25 ~ "5 (iii) Required probability = P(A \JB) = P(A) + P(B)-P(A C\B)
= 0.25 + 0.15-0.10=0.30==