Probability Exercise 01
Question:
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)
Answer:
(i) P(E∣F) = P(E∩F)/P(F) = 1/2
P(F∣E) = P(E∩F)/P(E) = 1/3
(ii) P(E∣G) = P(E∩G)/P(G) = 2/4 = 1/2
P(G∣E) = P(E∩G)/P(E) = 2/3
(iii) P((E∪F)∣G) = P((E∪F)∩G)/P(G) = 3/4
P((E∩F)∣G) = P((E∩F)∩G)/P(G) = 1/4
Question:
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.
A 0.33,0.11
B 0.51,0.76
C 0.56,0.43
D 0.11,0.65
Answer:
Answer: C 0.56,0.43
Question:
If A and B are events such that P(A∣B)=P(B∣A), then A A⊂B but A=B B A=B C A∩B=Φ D P(A)=P(B)
Answer:
A Incorrect B Correct C Incorrect D Incorrect
Question:
A coin is tossed three times, where
(i) E : head on third toss, F : heads on first two tosses
(ii) E : at least tow heads, F : at most two heads
(iii) E : at most two tails, F : at least one tail
Determine P(E∣F)
A 0.42,0.50,0.85
B 0.50,0.42,0.85
C 0.85,0.42,0.30
D 0.42,0.46,0.47
Answer:
Answer: D
Explanation:
Step 1: Calculate P(EF) for (i): P(EF) = P(E and F)/P(F) = P(head on third toss and heads on first two tosses)/P(heads on first two tosses) = 1/2 = 0.50
Step 2: Calculate P(EF) for (ii): P(EF) = P(E and F)/P(F) = P(at least two heads and at most two heads)/P(at most two heads) = 1/2 = 0.50
Step 3: Calculate P(EF) for (iii): P(EF) = P(E and F)/P(F) = P(at most two tails and at least one tail)/P(at least one tail) = 1/2 = 0.50
Therefore, the correct answer is D: 0.42, 0.46, 0.47.
Question:
If P(A)=1/2, P(B)=0, then P(A∣B) is A 0 B 1/2 C Not defined D 1
Answer:
Answer: C. Not defined
Question:
Given that E and F are events such that P(E)=0.6,P(F)=0.3 and P(E∩F)=0.2, find 6P(F∣E).
Answer:

P(F∣E) = P(E∩F) / P(E)

P(F∣E) = 0.2 / 0.6

P(F∣E) = 0.333

6P(F∣E) = 6 * 0.333

6P(F∣E) = 2
Question:
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’.
Answer:
Given that ‘at least one die shows a 3’,
P(Coin shows tail) = P(Coin shows tail  At least one die shows 3)
= P(Coin shows tail and At least one die shows 3) / P(At least one die shows 3)
= P(Coin shows tail and At least one die shows 3) / [P(One die shows 3) + P(Two dice show 3)]
= P(Coin shows tail) * P(At least one die shows 3  Coin shows tail) / [P(One die shows 3) + P(Two dice show 3)]
= 1/2 * [P(One die shows 3  Coin shows tail) + P(Two dice show 3  Coin shows tail)] / [P(One die shows 3) + P(Two dice show 3)]
= 1/2 * [1/2 + 1/4] / [1/2 + 1/4]
= 1/2
Question:
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 ?
Answer:
Answer: Step 1: Calculate the total number of questions in the question bank: 300 + 200 + 500 + 400 = 1400
Step 2: Calculate the total number of easy questions: 300 + 500 = 800
Step 3: Calculate the total number of multiple choice questions: 500 + 400 = 900
Step 4: Calculate the total number of easy multiple choice questions: 500
Step 5: Calculate the probability of selecting an easy multiple choice question: 500 / 1400 = 0.357
Question:
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 ? A 0.33,0.34 B 0.88,0.98 C 0.78,0.67 D 0.5,0.33
Answer:
Answer: B 0.88, 0.98
Explanation: (i) If the youngest is a girl, the probability that both are girls is 0.88. (ii) If at least one is a girl, the probability that both are girls is 0.98.
Question:
Compute 100×P(A∣B), if P(B)=0.5 and P(A∩B)=0.32.
Answer:
Step 1: P(A∩B) = P(A) × P(B)
Step 2: P(A) = P(A∩B) ÷ P(B)
Step 3: P(A) = 0.32 ÷ 0.5
Step 4: P(A) = 0.64
Step 5: 100 × P(A∣B) = 100 × 0.64
Step 6: 100 × P(A∣B) = 64
Question:
Evaluate P(A∪B), if 2P(A)=P(B)=5/13 and P(A∣B)=2/5. Find P(A∪B)×100
Answer:
Step 1: P(A∩B) = P(AB) × P(B) = (2/5) × (5/13) = 10/13
Step 2: P(A∪B) = P(A) + P(B)  P(A∩B) = (2P(A)) + P(B)  P(A∩B) = (2 × (5/13)) + (5/13)  (10/13) = 5/13
Step 3: P(A∪B) × 100 = (5/13) × 100 = 38.46%
Question:
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 Determine P(E∣F) A 1,0 B 1,0.2 C 0,2 D 1,3
Answer:
Answer: A) 1,0
Explanation: P(E∣F) is the probability of event E given event F has already occurred. Since event F states that one coin shows head, then event E must occur since it states that tail appears on one coin. Therefore, the probability of event E given event F is 1,0.
Question:
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'.
Answer:
Step 1: Calculate the total number of outcomes when two dice are thrown.
There are 36 possible outcomes when two dice are thrown.
Step 2: Calculate the number of favourable outcomes when the sum of numbers on the dice is 4.
There are 3 favourable outcomes when the sum of numbers on the dice is 4 (1+3, 2+2, 3+1).
Step 3: Calculate the probability of the event.
The probability of the event is 3/36 or 1/12.
Question:
If P(A)=116,P(B)=115 and P(A∪B)=117, find
(i) P(A∩B)
(ii) P(A∣B)
(iii) P(B∣A)
A 0.58,0.58,0.83
B 0.42,0.67,0.57
C 0.36,0.80,0.66
D 0.80,0.98,0.87
Answer:
(i) P(A∩B) = P(A) + P(B)  P(A∪B) = 116 + 115  117 = 114
(ii) P(A∣B) = P(A∩B) / P(B) = 114 / 115 = 0.98
(iii) P(B∣A) = P(A∩B) / P(A) = 114 / 116 = 0.98
Hence, the answer is D 0.80,0.98,0.87.
Question:
Mother, father and son line up at random for a family picture E: son on one end,
Answer:
A: First, the son should stand on one end of the line. B: Then, the father should stand in the middle of the line. C: Finally, the mother should stand on the other end of the line.
Question:
F: father in middle. Determine P(E∣F)
Answer:
Step 1: Identify the given statement. The given statement is a conditional probability statement. It states that the probability (P) of event E given that event F has occurred.
Step 2: Define the events E and F. Event E: Father is in the middle Event F: Father is present
Step 3: Calculate the probability of event E given event F. P(E∣F) = P(E ∩ F) / P(F) P(E ∩ F) = 1 since when father is present, he must be in the middle P(F) = 1 since father is present Therefore, P(E∣F) = 1/1 = 1
Question:
A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses. Determine 36P where P(E∣F)
Answer:
Step 1: P(E∣F) = P(E∩F)/P(F)
Step 2: P(E∩F) = P(E) x P(F)
Step 3: P(E) = 1/6, P(F) = (1/6 x 1/6) = 1/36
Step 4: P(E∩F) = (1/6 x 1/36) = 1/216
Step 5: P(E∣F) = 1/216/1/36 = 1/6
Question:
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)
Answer:
(i) P(A∩B) = P(B∣A)P(A) = 0.4 × 0.8 = 0.32
(ii) P(A∣B) = P(A∩B)/P(B) = 0.32/0.5 = 0.64
(iii) P(A∪B) = P(A) + P(B)  P(A∩B) = 0.8 + 0.5  0.32 = 1.18