Referring to the AP Inter 2nd Year Maths Study Material Chapter 13 Probability Exercise 13a Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Probability Solutions Exercise 13a
I.
Question 1.
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P (F|E)
Solution:
Given that P(E) = 0.6, P(F) = 0.3, P(E ∩ F) = 0.2
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}\) and
P(F/E) = \(\frac{P(F \cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\)
Question 2.
Compute P(A|B), if P(B) = 0.5 and P(A ∩ B) = 0.32
Solution:
Given that P(B) = 0.5 and P(A ∩ B) = 0.32
∴ P(A|B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
= \(\frac{0.32}{0.5}\) = \(\frac{32}{50}\)
= \(\frac{16}{25}\)
Question 3.
A and B are two events such that P (A) ≠ 0. Find P(B|A), if
(i) A is a subset of B
(ii) A ∩ B = Φ
Solution:
Given,P(A) ≠ 0
(i) A is a subset of B ⇒ A ∩ B = A – Then P(A ∩ B) = P(B ∩ A) = P(A)
∴P(B|A) = \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 1
(ii) A ∩ B Φ ⇒ P (A ∩ B) = 0
∴ P(B|A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\) = 0
Question 4.
If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
Solution:
In a leap year, there are 366 days i.e., 52 weeks and 2 days.
In 52 weeks, there are 52 Tuesdays.
∴ the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining 2 days will be Tuesdays.
The remaining 2 days can be any of the following:
Monday and Tuesday or Tuesday and Wednesday or Wednesday and Thursday or Thursday and Friday or Friday and Saturday or Saturday and Sunday or Sunday and Monday.
Total number of cases = 7
Favourable cases = 2
Probability that a leap year will have 53 Tuesdays = 2/7
II.
Question 1.
If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(A ∪ B)
Solution:
Given that P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4
(i) Now P (B|A) = 0.4 ⇒ \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 0.4
⇒ \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{0.8}\) = 0.4
⇒ P(A ∩ B) = 0.8 × 0.4 = 0.32
(ii) P(A|B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.5}\)
= \(\frac{32}{100} \times \frac{10}{5}\) = \(\frac{64}{100}\) = 0.64
(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.8 + 0.5 – 0.32 = 1.3- 0.32 = 0.98.
Question 2.
Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A | B) = \(\frac{5}{5}\)
Solution:
Given that 2P(A)= P(B) = \(\frac{5}{13}\) =>P(A) = \(\frac{5}{26}\), P(B) = \(\frac{5}{13}\)
Now P(A|B) = \(\frac{2}{5}\) ⇒ \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2}{5}\)
⇒ P(A ∩ B) = \(\frac{2}{5}\)P(B) = \(\frac{2}{5}\) × \(\frac{5}{13}\) = \(\frac{2}{13}\)
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{5}{26}\) + \(\frac{5}{13}\) – \(\frac{2}{13}\) = \(\frac{5+10-4}{26}\)
= \(\frac{11}{26}\)
Question 3.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) andP(A ∪ B) = \(\frac{7}{11}\), find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(B|A)
Solution:
(i) Given that P(A ∪ B) = \(\frac{7}{11}\)
⇒ P(A) + P(B) – P(A ∩ B) = \(\frac{7}{11}\)
⇒ \(\frac{6}{11}\) + \(\frac{5}{11}\) – P(A ∩ B) = \(\frac{7}{11}\)
⇒ P(A ∩ B) = 1 – \(\frac{7}{11}\)= \(\frac{4}{11}\)
(ii) P(A/B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
= \(\frac{\frac{4}{11}}{\frac{5}{11}}\) = \(\frac{4}{5}\)
(iii) P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\)
= \(\frac{\frac{4}{11}}{\frac{6}{11}}\) = \(\frac{2}{3}\)
Question 4.
Determine P(E|F) when a coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
Solution:
We know that the sample space for the random experiment ‘a coin is tossed three times’ is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} ⇒ n(S) = 8
(i) E: Head on third toss ⇒ E = {HHH, HTH, THH, TTH} ⇒ n(E) = 4
F: Heads on first two tosses ⇒ F = {HHH, HHT} ⇒ n(F) = 2
Hence E ∩ F = {HHH} ⇒ n(E ∩ T) = 1
Now P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)
P(F) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)
P(E ∩ F) = \(\frac{1}{8}\)
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{\frac{1}{8}}{\frac{1}{4}}\) = \(\frac{1}{2}\)
Question 5.
Determine P(E|F) when a coin Is tossed three times, where
E : at least two heads , F: at most two heads
Solution:
E : at least two heads ⇒ E = {HHII, HHT, HTH, THH} = n(E) = 4
F: at most two beads ⇒ F = {HI-IT, HTH, THH, HTT, THT, TTH, TTT} ⇒ n(F) 7
Hence E ∩ F = {HHT, HTH, THH} ⇒ n(E ∩ F) = 3
Now P(E) = \(\frac{4}{8}\) = \(\frac{1}{2}\) ,
P(F) = \(\frac{7}{8}\),
P(E ∩ F) = \(\frac{\mathrm{n}(\mathrm{E} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{8}\)
∴ P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}\)
= \(\frac{\frac{3}{8}}{\frac{7}{8}}\) = \(\frac{3}{7}\)
Question 6.
Determine P(EIF) when a coin is tossed three times, where
E : at most two tails , F : at least one tail.
Solution:
E : at moat two tails ⇒ E = {TTH, THT, HTT, THH, HTH, HHT, HHH} ∴ n(E) = 7
F: at least one tail ⇒ F = {THH, HTH, HHT, TTH, THT, HTT, TTT) ∴ n(F) = 7
Hence E ∩ F = {TTH, THT, HTT, THH, HTH, HHT} ⇒ n(E ∩ F) = 6
Now P(E) = \(\frac{7}{8}\), P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{7}{8}\),
P(E ∩ F) = \(\frac{6}{8}\) = \(\frac{3}{4}\)
∴ P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}\) = \(\frac{\frac{3}{4}}{\frac{7}{8}}\)
= \(\frac{3}{4} \times \frac{8}{7}\) = \(\frac{6}{7}\)
Question 7.
Find P(E|F) when two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E : no tail appears, F : no head appears.
Solution:
When ‘two coins are tossed once’ in the sample space S = {HH, HT, TH, TT} ⇒ n(S) = 4
(i) E : tail appears on one coin ⇒ E = {HT, TH} ⇒ n(E) = 2
F : one coin shows head ⇒ F = {HT, TH} ⇒ n(F) = 2
Hence E ∩ F = {HT, TH} ⇒ n(E ∩ F) = 2
Now P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(F) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(E ∩ F) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{\frac{1}{2}}{\frac{1}{2}}\) = 1
(ii) E: no tail = {HH}; F: no head = {TT}
∴ E ∩ F = Φ; P(F) = 1 and P (E ∩ F) = 0
∴ P(E|F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{0}{1}\) = 0
Question 8.
Find P(E|F) when a die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively on first two tosses.
Solution:
When a die is thrown 3 times, n(S) = 63 = 216
E: 4 appears on third toss = {(1, 1, 4) (1, 2, 4)… (1, 6, 4) (2, 1, 4) (2, 2, 4)…
(2, 6, 4)(3, 1, 4) (3, 2, 4) … (3 6 4) (4, 1, 4)(4, 2, 4)… (4, 6, 4)(5, 1, 4) (5, 2, 4)… (5, 6, 4) (6, 1, 4) (6, 2, 4)…
F: 6 and 5 appear respectively on first two tossess
= {(6, 5, 1) (6, 5, 2) (6, 5, 3) (6, 5, 4) (6, 5, 5) (6, 5, 6)}
Hence E ∩ F = {(6, 5, 4)}
Now P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)
∴ P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\left(\frac{1}{2 \times 6}\right)}{\left(\frac{6}{2 \times 6}\right)}\)
= \(\frac{1}{6}\)
Question 9.
Determine P(E|F) when Mother, father and son line up at random for a family picture E : son on one end, F : father in middle. Find P(E|F)
Solution:
Let m, f and s denote the mother, father and son respectively.
The sample space is S = {mfs, msf, fms, fsm, smf, sfm} ∴ n(S) = 6
E: son on one end ⇒ E = {mfs, fins, smf, sftn} ⇒ n(E) = 4 ⇒ P(E) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
F: father in middle F= {mfs, sfm} ⇒ n(F) = 2 ⇒ P(F) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Hence, E ∩ F = {mfs, sfm} ⇒ n(E ∩ F) = 2 ⇒ P(E ∩ F) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
∴ P(E|F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{1 / 3}{1 / 3}\) = 1.
Question 10.
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl, (ii) at least one is a girl?
Solution:
Let the first (elder) child be denoted by capital letter and the second (younger) by a small
letter. The sample space is S = {Bb, Bg, Gb, Gg} ⇒ n(S) = 4
Let E: both children are girls, then E = {Gg.} ⇒ n(E) = 1 ⇒ P(E) = \(\frac{1}{4}\)
(i) Let F: the youngest child is a girl, then
F = {Bg, Gg} ⇒ n(F) = 2 ⇒ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{4}\)
Hence E ∩ F = {Gg} ⇒ n(E ∩ F) = 1 ⇒ P(E ∩ F) = \(\frac{1}{4}\)
∴ P(E|F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{4}}{\frac{2}{4}}\) = \(\frac{1}{2}\)
(ii) Let F : at least one child is a girl then F {Bg, Gb, Gg} ⇒ n(F) = 3
⇒ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{4}\)
Hence, E ∩ F = {Gg}:. n(E ∩ F) = 1 ⇒ P(E ∩ F) = \(\frac{1}{4}\)
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}\) = \(\frac{1}{3}\)
Question 11.
An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions- and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution:
Total number of questions = 300 + 200 + 500 + 400 = 1400
∴ n(S) = 1400
Let E : selected question is easy
F : selected question is a multiple choice question then
E ∩ F : selected question is an easy multiple choice question
Thus n(E ∩ F) = 500, n(F) = 500 + 400 = 900
∴ P(E ∩ F) = \(\frac{n(\mathrm{E} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{500}{1400}\) and P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{900}{1400}\)
∴ Required probability is P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\)
= \(\frac{\frac{500}{1400}}{\frac{900}{1400}}\) = \(\frac{500}{1400} \times \frac{1400}{900}\)
= \(\frac{5}{9}\)
Question 12.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:
Sample space of throwing two dice is
S = {( x, y): x, y e {1, 2, 3, 4, 5, 6}} ⇒ n(S) = 6 × 6 = 36
Let E : the sum of numbers on the dice is 4 ⇒ E = {(1, 3), (2, 2), (3,1)}
⇒ n(E) = 3 ⇒ P(E) = \(\frac{n(E)}{n(S)}=\frac{3}{36}\)
Let F : numbers appearing on the dice are different
⇒ F = S – {(1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6)}
∴ n(F) = 36 – 6 = 30 ⇒ P(F) = \(\frac{30}{36}\)
Also, E ∩ F = {(1,3), (3,1)} ⇒ n(E ∩ F) = 2
∴ P(E ∩ F) = \(\frac{n(E \cap F)}{n(S)}=\frac{2}{36}\)
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{\frac{2}{36}}{\frac{30}{36}}=\) = \(\frac{2}{36} \times \frac{36}{30}\)
= \(\frac{1}{15}\)
Question 13.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Solution:
The sample space of given random experiment is
S = {(1, H), (1, T), (2, H), (2, T), (3,1), (3,2), (3,3), (3, 4) (3, 5), (3, 6), (4, H), (4, T), (5, H), (5, T), (6, 1), (6, 2),(6, 3), (6, 4), (6, 5), (6, 6)}
⇒ n(S) = 20
Let E: the coin shows a tail ⇒ E= {(1, T), (2, T), (4, T), (5, T)} ⇒ n(E) = 4
Let F: at least one die shows a 3
⇒ F = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)} ⇒ n(F) = 7
Hence, E ∩ F = Φ ⇒ n(E ∩ F) = 6 ⇒ P(E ∩ F) = \(\frac{n(E \cap F)}{n(S)}=\frac{0}{20}\) = 0.
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0}{P(F)}\) = 0.
Question 14.
A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
Solution:
If a couple has two children, then the sample space is S = {(B,B), (B,G), (QB), (G,G)}
(i) Let E and F respectively denote the events that both children are male and atleast onechildren is a male. .
E ∩ F ={(G,G)} ⇒ P(E ∩ F) = \(\frac{1}{4}\)
P(E) = \(\frac{1}{4}\)
P(F) = \(\frac{3}{4}\)
⇒ P(E|F) = \(\frac{P(E \cap F)}{P(F)}\) = \(\frac{1 / 4}{3 / 4}=\frac{1}{4} \times \frac{4}{3}\) = \(\frac{1}{3}\)
(ii) Let C and D respectively denote the events that both children are females and the elder child is a female.
C = {(G, G)} ⇒ P(C) = \(\frac{1}{4}\); D = {(G, B),(G, G)} ⇒ P(D) = \(\frac{2}{4}\)
C ∩ D = {(G, G)} ⇒ P(C ∩ D) = \(\frac{1}{4}\)
∴ P(C|D) = \(\frac{\mathrm{P}(\mathrm{C} \cap \mathrm{D})}{\mathrm{P}(\mathrm{D})}=\frac{1 / 4}{2 / 4}\) = \(\frac{1}{2}\)
Question 15.
If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability \(\frac{1}{2}\)).
Solution:
The total number of determinants of second order with each element being 0 or 1 is (2)4 = 16.
The value of determinant is positive in the following cases. \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|,\left|\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right|,\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|\)
∴ Required probability = \(\frac{3}{16}\)
Question 16.
An electronic assembly consists of two sub systems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15
Evaluate the following probabilities (i) P(A fails|B has failed)
(ii) P(A fails alone)
Solution:
Let the event in which A fails and B fails he denote by EA and EB
P(EA) = 0.2,P(EA ∩ EB) = 0.15 .
P(B fails alone) = P(EB) – P(EA ∩ EB)
∴ 0.15 = P(EB) – 0.15
∴ P(EB) = 0.3
(i) P(EA |EB) = \(\frac{P\left(E_A \cap E_B\right)}{P\left(E_B\right)}\) = \(\frac{0.15}{0.3}\) = 0.5
(ii) P(A fails alone) = P(EA) – P(EA and EB) = 0.2 – 0.15 = 0.05
III.
Question 1.
A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:
Let x denote the outcome on black die and y denote the outcome on red die.
The sample space is S = {(x, y) : x, y ∈ {1, 2, 3, 4, 5, 6}} ⇒ n(S) = 6 × 6 = 36
(a) LetE : sum x + y > 9 ⇒ x + y = 10, 11, 12
⇒ E = {(6,4), (6,5), (6,6), (5, 5), (5,6), (4,6)}
F : black die resulted in a 5. ⇒ F = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ n(F) = 6 ⇒ P(F) = \(\frac{6}{36}\)
Hence, E ∩ F = {(5, 5), (5, 6)}
∴ n(E ∩ F) = 2 ⇒ P(E ∩ F) = \(\frac{2}{36}\)
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{6}{36}}\) = \(\frac{6}{3}\)
(b) Let E : sum x + y = 8 ⇒ E = {(2, 6), (3, 5), (4,4), (5, 3), (6, 2)}
F : red die resulted in a number less than 4
⇒ F = {(x, y) : x ∈ {1, 2, 3, 4, 5, 6} and y ∈ {1, 2, 3}}
= {(1,1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1),(3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
∴ n(F) = 6 × 3 = 18 ⇒ P(F) = \(\frac{18}{36}\)
Also, E ∩ F = {(5, 3), (6, 2)}
∴ n(E ∩ F) = 2 ⇒ P(E ∩ F) ⇒ P(E ∩ F) = \(\frac{2}{36}\)
∴ Required probability is P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{18}{36}}\) = \(\frac{1}{9}\)
Question 2.
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find
(i) P (E|F) and P (F|E)
(ii) P (E|G) and P(G|E)
(iii) P ((E ∪ F)|G) and P ((E ∩ F)|G)
Solution:
Sample space S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
Given: Event E = {1, 3, 5}, F = {2, 3}, G = {2, 3, 4, 5}
(i) E ∩ F = {3}, n(E) = 3, n(F) = 2, n(G) = 4, n(E ∩ F) = 1
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{3}{6}\), P(F) = \(\frac{2}{6}\), P(G) = \(\frac{4}{6}\), P(E ∩ F) = \(\frac{1}{6}\)
∴ P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{6}}{\frac{2}{6}}\) = \(\frac{1}{2}\) and P(F|E) = \(\frac{P(F \cap E)}{P(E)}=\frac{\frac{1}{6}}{\frac{3}{6}}\) = \(\frac{1}{3}\)
(ii) Now E ∩ G = {3, 5} ⇒ n(E ∩ G) = 2 ⇒ P(E ∩ G) = \(\frac{2}{6}\)
∴ P(E|G) = \(\frac{P(E \cap G)}{P(G)}=\frac{\frac{2}{6}}{\frac{4}{6}}\) = \(\frac{1}{2}\) and P(G|E) = \(\frac{P(E \cap G)}{P(E)}=\frac{\frac{2}{6}}{\frac{3}{6}}\) = \(\frac{2}{3}\)
(iii) Also E ∪ F = {1, 2, 3, 5}
⇒ E ∩ F = {3} (E ∪ F) ∩ G = {1, 2, 3, 5} ∩ {2, 3, 4, 5} = {2, 3, 5}
⇒ n((E ∪ F) ∩ G) = 3 ⇒ P((E ∪ F) ∩ G) = \(\frac{3}{6}\)
Now (E ∩ F) ∩ G = {3} ∩ {2, 3, 4, 5} = {3}
⇒ n((E ∩ F) ∩ G) = 1 ⇒ P((E ∩ F) ∩ G) = \(\frac{1}{6}\)
∴ P((E ∪ F)|G) = \(\frac{P((E \cup F) \cap G)}{P(G)}=\frac{\frac{3}{6}}{\frac{4}{6}}\) = \(\frac{3}{4}\)
P((E ∩ F)|G) = \(\frac{P((E \cap F) \cap G)}{P(G)}=\frac{\frac{1}{6}}{\frac{4}{6}}\) = \(\frac{1}{4}\)