Probability Exercise 03
Question:
Which of the following can not be valid assignment of probabilities for outcomes of sample space S = {{ω1,ω2,ω3,ω4,ω5,ω6,ω7} Assignment ω1 ω2 ω3 ω4 ω5 ω6 ω7 (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 (b) 71 71 71 71 71 71 71 (c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (d) 0.1 0.2 0.3 0.4 0.2 0.1 0.3 (e) 141 142 143 144 145 146 1415
Answer:
Answer: (b) 71 71 71 71 71 71 71
(d) 0.1 0.2 0.3 0.4 0.2 0.1 0.3
These two cannot be valid assignments of probabilities for outcomes of sample space S.
Question:
A die is thrown find the probability of following events: (i) A prime number will appear (ii) A number greater than or equal to 3 will appear (iii) A number less than or equal to one will appear (iv) A number more than 6 will appear (v) A number less than 6 will appear
Answer:
(i) A prime number will appear: Since a die has six faces numbered 1, 2, 3, 4, 5, and 6, the probability of a prime number appearing is 2/6 or 1/3.
(ii) A number greater than or equal to 3 will appear: Since a die has six faces numbered 1, 2, 3, 4, 5, and 6, the probability of a number greater than or equal to 3 appearing is 4/6 or 2/3.
(iii) A number less than or equal to one will appear: Since a die has six faces numbered 1, 2, 3, 4, 5, and 6, the probability of a number less than or equal to one appearing is 1/6.
(iv) A number more than 6 will appear: Since a die has six faces numbered 1, 2, 3, 4, 5, and 6, the probability of a number more than 6 appearing is 0/6 or 0.
(v) A number less than 6 will appear: Since a die has six faces numbered 1, 2, 3, 4, 5, and 6, the probability of a number less than 6 appearing is 5/6.
Question:
There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Answer:

There are a total of 10 council members (4 men and 6 women).

The probability of selecting a woman is 6/10, or 3/5.
Question:
Events E and F are such that P(not E or not F) =0.25 State whether E and F are mutually exclusive
Answer:

First, calculate P(E and F). This can be done using the formula P(not E or not F) = 1  P(E and F).

P(not E or not F) = 0.25

Therefore, P(E and F) = 1  0.25 = 0.75

Since P(E and F) is not equal to 0, this means that E and F are not mutually exclusive.
Question:
In an entrance test is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least of them is 0.95. What is the probability of passing both
Answer:
P(pass both) = P(pass first) * P(pass second)
P(pass both) = 0.8 * 0.7
P(pass both) = 0.56
Question:
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75 What is the probability of passing the Hindi examination
Answer:
P(passing English) = 0.75
P(passing Hindi) = P(passing both) + P(passing Hindi only) = 0.5 + P(passing Hindi only)
P(passing neither) = 0.1
P(passing Hindi only) = P(passing neither)  P(passing both) = 0.1  0.5 = 0.4
Therefore, P(passing Hindi) = 0.5  0.4 = 0.1
Question:
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS (ii) The student has opted neither NCC nor NSS (iii) The student has opted NSS but not NCC
Answer:
(i) P(NCC or NSS) = P(NCC) + P(NSS)  P(NCC and NSS) = 30/60 + 32/60  24/60 = 38/60
(ii) P(neither NCC nor NSS) = 1  P(NCC or NSS) = 1  38/60 = 22/60
(iii) P(NSS but not NCC) = P(NSS)  P(NCC and NSS) = 32/60  24/60 = 8/60
Question:
Given, P(A)=3/5 and P(B)=1/5. Find P(AorB), if A and B are mutually exclusive events.
Answer:
Answer: Step 1: A and B are mutually exclusive events, which means that they cannot occur together.
Step 2: Since they cannot occur together, P(A or B) = P(A) + P(B)
Step 3: P(A or B) = 3/5 + 1/5 = 4/5
Question:
A coin is tossed twice what is the probability that at least one tail occurs?
Answer:
Step 1: Calculate the number of possible outcomes when a coin is tossed twice.
There are four possible outcomes when a coin is tossed twice: TT (two tails), TH (one tail, one head), HT (one head, one tail), and HH (two heads).
Step 2: Calculate the number of outcomes where at least one tail occurs.
There are three outcomes where at least one tail occurs: TT, TH, and HT.
Step 3: Calculate the probability of at least one tail occurring.
The probability of at least one tail occurring is 3/4.
Question:
Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A)=0.5,P(B)=0.7, P(A∩B)=0.6 (ii) P(A)=0.5,P(B)=0.4, P(A∪B)=0.8
Answer:
(i) Yes, the probabilities are consistently defined.
P(A∩B) = P(A) × P(B) = 0.5 × 0.7 = 0.35
(ii) No, the probabilities are not consistently defined.
P(A∪B) = P(A) + P(B)  P(A∩B) = 0.5 + 0.4  P(A∩B) ≠ 0.8
Question:
A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12
Answer:
(i) P(sum = 3) = P(Coin = 1, Die = 2) + P(Coin = 6, Die = 2) = (1/2)(1/6) + (1/2)(1/6) = 1/3
(ii) P(sum = 12) = P(Coin = 1, Die = 11) + P(Coin = 6, Die = 6) = (1/2)(1/6) + (1/2)(1/6) = 1/3
Question:
Three coins are tossed once Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails
Answer:
(i) P(3 heads) = 1/8
(ii) P(2 heads) = 3/8
(iii) P(at least 2 heads) = 4/8
(iv) P(at most 2 heads) = 7/8
(v) P(no head) = 1/8
(vi) P(3 tails) = 1/8
(vii) P(exactly two tails) = 3/8
(viii) P(no tail) = 1/8
(ix) P(at most two tails) = 7/8
Question:
If 2/11 is the probability of an event. What is the probability of the event ’not A'
Answer:
Answer:
 The probability of an event ’not A’ is 1  2/11
 Therefore, the probability of the event ’not A’ is 9/11
Question:
A letter is chosen at random from the word ′ASSASSINATION′. Find the probability that letter is (i) a vowel (ii) an consonant
Answer:
(i) Probability of choosing a vowel = 4/10 = 2/5
(ii) Probability of choosing a consonant = 6/10 = 3/5
Question:
A card is selected from a pack of 52 cards (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades (c) Calculate the probability that the card is (i) an ace (ii) black card
Answer:
(a) The sample space consists of 52 points as there are 52 cards in the pack.
(b) The probability that the card is an ace of spades is 1/52 as there is only one ace of spades in the pack.
(c) The probability that the card is an ace is 4/52 as there are 4 aces in the pack. The probability that the card is a black card is 26/52 as there are 26 black cards in the pack.
Question:
A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs. 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Answer:
Step 1: The sample space for a fair coin being tossed four times can be represented as {HHHT, HTHH, THHH, HHTH, THTH, HTHT, TTHH, HTTH}.
Step 2: The possible amounts of money that you can have after four tosses are: Re 1, Re 0.50, Re 0, Re 1.50, Re 2, Re 1.
Step 3: The probability of having each of these amounts can be calculated by counting the number of outcomes that result in each amount. For example, the probability of having Re 1 is 2/8, as there are two outcomes (HHHT and HTHH) that result in Re 1. Similarly, the probability of having Re 1.50 is 2/8, as there are two outcomes (THTH and HTTH) that result in Re 1.50.
Step 4: The probabilities of having each of the possible amounts of money after four tosses are as follows: Re 1 (2/8), Re 0.50 (2/8), Re 0 (2/8), Re 1.50 (2/8), Re 2 (1/8), Re 1 (1/8).
Question:
In a lottery, a person choses six different natural numbers at random from 1 to 20 and if these six numbers match with the six numbers already fixed by the lottery committee he wins the prize. What is the probability of winning the prize in the game? [Hint : order of the numbers is not important]
Answer:
Step 1: Calculate the total number of possible combinations of 6 numbers that can be chosen from 1 to 20. This can be done by using the combination formula, nCr, where n = 20 and r = 6.
Step 2: Calculate the total number of combinations that match the numbers already fixed by the lottery committee. This can be done by counting the number of combinations that match the numbers.
Step 3: Calculate the probability of winning the prize by dividing the number of combinations that match the numbers by the total number of possible combinations.
Probability of winning the prize = Number of combinations that match the numbers / Total number of possible combinations
Question:
Fill in the blanks in following table: P(A) P(B) P(A∩B) P(A∪B) (i) 1/3 1/5 1/15 …… (ii) 0.35 ….. 0.25 0.6 (iii) 0.5 0.35 …. 0.7
Answer:
(i) 1/3 1/5 1/15 2/15 (ii) 0.35 0.2 0.25 0.6 (iii) 0.5 0.35 0.35 0.7
Question:
If E and F are events such that P(E)= 1/4, P(F)= 1/2 and P(EandF)= 1/8 find: (i) P(E or F) (ii) P(not E and not F)
Answer:
(i) P(E or F) = P(E) + P(F)  P(E and F) = 1/4 + 1/2  1/8 = 7/8
(ii) P(not E and not F) = 1  P(E or F) = 1  7/8 = 1/8
Question:
A and B are events such that P(A)=0.42,P(B)=0.48 and P(A and B) =0.16 Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)
Answer:
(i) P(not A) = 1  P(A) = 1  0.42 = 0.58
(ii) P(not B) = 1  P(B) = 1  0.48 = 0.52
(iii) P(A or B) = P(A) + P(B)  P(A and B) = 0.42 + 0.48  0.16 = 0.74