Chapter 16 Probability Ex 16.3 Solutions
Question - 1 : - Which of the following cannot bevalid assignment of probabilities for outcomes of sample Space S = {ω1,ω2, ω3, ω4, ω5, ω6, ω7}
Assignment
| Assignment | ω1 | ω2 | ω3 | ω4 | ω5 | ω6 | ω7 |
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 |
| (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) | 1/14 | 2/14 | 3/14 | 4/14 | 5/14 | 6/14 | 15/14 |
Answer - 1 : -
(a) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Therefore, the given assignment is valid.
b) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
= (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7)
= 7/7
= 1
Therefore, the given assignment is valid.
c) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7
= 2.8 > 1
Therefore, the 2nd condition is notsatisfied
Which states that p(wi) ≤ 1
So, the given assignment is not valid.
d) The conditions of axiomatic approach don’t hold true inthe given assignment, that is
1) Each of the number p(wi) is less than zero butalso negative
To be true each of the number p(wi) should beless than zero and positive
So, the assignment is not valid
e) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
= (1/14) + (2/14) + (3/14) + (4/14) + (5/14) + (6/14) +(7/14)
= (28/14) ≥ 1
The second condition doesn’t hold true so the assignment isnot valid.
Question - 2 : - A coin is tossed twice, what is the probability that at least one tail occurs?
Answer - 2 : -
Since either coin can turn up Head (H) or Tail (T), are thepossible outcomes.
Here coin is tossed twice, then sample space is S = (TT, HH,TH, HT)
∴ Number of possible outcomes n (S) = 4
Let A be the event of getting at least one tail
∴ n (A) = 3
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
P(A) = n(A)/n(S)
= ¾
Question - 3 : - 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 - 3 : -
Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
Here S = {1, 2, 3, 4, 5, 6}
∴n(S) = 6
(i) A prime number will appear,
Let us assume ‘A’ be the event of getting a prime number,
A = {2, 3, 5}
Then, n(A) = 3
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(A) = n(A)/n(S)
= 3/6
= ½
(ii) A number greater than or equal to 3 will appear,
Let us assume ‘B’ be the event of getting a number greaterthan or equal to 3,
B = {3, 4, 5, 6}
Then, n(B) = 4
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(B) = n(B)/n(S)
= 4/6
= 2/3
(iii) A number less than or equal to one will appear,
Let us assume ‘C’ be the event of getting a number less thanor equal to 1,
C = {1}
Then, n (C) = 1
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(C) = n(C)/n(S)
= 1/6
(iv) A number more than 6 will appear,
Let us assume ‘D’ be the event of getting a number more than6, then
D = {0)}
Then, n (D) = 0
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(D) = n(D)/n(S)
= 0/6
= 0
(v) A number less than 6 will appear.
Let us assume ‘E’ be the event of getting a number less than6, then
E= (1, 2, 3, 4, 5)
Then, n (E) = 5
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(E) = n(E)/n(S)
= 5/6
Question - 4 : - 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 - 4 : -
From the question it is given that, there are 52 cards inthe deck.
(a) Number of points in the sample space = 52 (given)
∴n(S) = 52
(b) Let us assume ‘A’ be the event of drawing an ace ofspades.
A= 1
Then, n (A) = 1
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(A) = n(A)/n(S)
= 1/52
(c) Let us assume ‘B’ be the event of drawing an ace. Thereare four aces.
Then, n (B)= 4
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(B) = n(B)/n(S)
= 4/52
= 1/13
(d) Let us assume ‘C’ be the event of drawing a black card.There are 26 black cards.
Then, n (C) = 26
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(C) = n(C)/n(S)
= 26/52
= ½
Question - 5 : - 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 - 5 : -
Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
So, the sample space S = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5),(6, 6)}
Then, n(S) = 12
(i) Let us assume ‘P’ be the event having sum of numbers as3.
P = {(1, 2)},
Then, n (P) = 1
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(P) = n(P)/n(S)
= 1/12
(ii) Let us assume ‘Q’ be the event having sum of number as12.
Then Q = {(6, 6)}, n (Q) = 1
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(Q) = n(Q)/n(S)
= 1/12
Question - 6 : - 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 - 6 : -
From the question it is given that, there are four men andsix women on the city council.
Here total members in the council = 4 + 6 = 10,
Hence, the sample space has 10 points
∴ n (S) = 10
Number of women are 6 … [given]
Let us assume ‘A’ be the event of selecting a woman
Then n (A) = 6
P(Event) = Number of outcomes favourable to event/Totalnumber of possible outcomes
∴P(A) = n(A)/n(S)
= 6/10 … [divide both numerator and denominators by 2]
= 3/5
Question - 7 : - A fair coin is tossed four times, and a person win Rs 1 for each head and lose 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 - 7 : -
Since either coin can turn up Head (H) or Tail (T), are thepossible outcomes.
But, now coin is tossed four times so the possible samplespace contains,
S = (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH,THTH, TTHH,
TTTH, TTHT, THTT, HTTT, TTTT)
As per the condition given the question, a person will winor lose money depending up on the face of the coin so,
(i) For 4 heads = 1 + 1 + 1 + 1 = ₹ 4
So, he wins ₹ 4
(ii) For 3 heads and 1 tail = 1 + 1 + 1 – 1.50
= 3 – 1.50
= ₹ 1.50
So, he will be winning ₹ 1.50
(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50
= 2 – 3
= – ₹ 1
So, he will be losing ₹ 1
(iv)For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50
= 1 – 4.50
= – ₹ 3.50
So, he will be losing Rs. 3.50
(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50
= – ₹ 6
So, he will be losing Rs. 6
Now the sample space of amounts is
S= {4, 1.50, 1.50, 1.50, 1.50, – 1, – 1, – 1, – 1, – 1, – 1,– 3.50, – 3.50, – 3.50, – 3.50, – 6}
Then, n (S) = 16
P (winning ₹ 4) = 1/16
P (winning ₹ 1.50) = 4/16 … [divide both numerator anddenominator by 4]
= ¼
P (winning ₹ 1) = 6/16 … [divide both numerator anddenominator by 2]
= 3/8
P (winning ₹ 3.50) = 4/16 … [divide both numerator anddenominator by 4]
= ¼
P (winning ₹ 6) = 1/16
= 3/8
Question - 8 : - 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 - 8 : -
Since either coin can turn up Head (H) or Tail (T), are thepossible outcomes.
But, now three coin is tossed so the possible sample spacecontains,
S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
(i) 3 heads
Let us assume ‘A’ be the event of getting 3 heads
n(A)= 1
∴P(A) = n(A)/n(S)
= 1/8
(ii) 2 heads
Let us assume ‘B’ be the event of getting 2 heads
n (A) = 3
∴P(B) = n(B)/n(S)
= 3/8
(iii) at least 2 heads
Let us assume ‘C’ be the event of getting at least 2 head
n(C) = 4
∴P(C) = n(C)/n(S)
= 4/8
= ½
(iv) at most 2 heads
Let us assume ‘D’ be the event of getting at most 2 heads
n(D) = 7
∴P(D) = n(D)/n(S)
= 7/8
(v) no head
Let us assume ‘E’ be the event of getting no heads
n(E) = 1
∴P(E) = n(E)/n(S)
= 1/8
(vi) 3 tails
Let us assume ‘F’ be the event of getting 3 tails
n(F) = 1
∴P(F) = n(F)/n(S)
= 1/8
(vii) Exactly two tails
Let us assume ‘G’ be the event of getting exactly 2 tails
n(G) = 3
∴P(G) = n(G)/n(S)
= 3/8
(viii) no tail
Let us assume ‘H’ be the event of getting no tails
n(H) = 1
∴P(H) = n(H)/n(S)
= 1/8
(ix) at most two tails
Let us assume ‘I’ be the event of getting at most 2 tails
n(I) = 7
∴P(I) = n(I)/n(S)
= 7/8
Question - 9 : - If 2/11 is the probability of an event, what is the probability of the event ‘not A’.
Answer - 9 : -
From the question it is given that, 2/11 is the probabilityof an event A,
i.e. P (A) = 2/11
Then,
P (not A) = 1 – P (A)
= 1 – (2/11)
= (11 – 2)/11
= 9/11
Question - 10 : - A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) a consonant
Answer - 10 : -
The word given in the question is ‘ASSASSINATION’.
Total letters in the given word = 13
Number of vowels in the given word = 6
Number of consonants in the given word = 7
Then, the sample space n(S) = 13
(i) a vowel
Let us assume ‘A’ be the event of selecting a vowel
n(A) = 6
∴P(A) = n(A)/n(S)
= 6/13
(ii) Let us assume ‘B’ be the event of selecting theconsonant
n(B)= 7
∴P(B) = n(B)/n(S)
= 7/13