Chapter 16: Probability Exercise 16.3

Solution:-

(a) Condition (i): Each of the numbers p(ω_i ) 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 numbers p(ω_i ) 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 numbers p(ω_i ) 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 2^nd condition is not satisfied.

Which states that p(w_i) ≤ 1

So, the given assignment is not valid.

d) The conditions of the axiomatic approach don’t hold true in the given assignment, that is

1) Each of the numbers p(w_i) is less than zero but also negative.

To be true, each of the numbers p(w_i) should be less than zero and positive.

So, the assignment is not valid.

e) Condition (i): Each of the numbers p(ω_i ) 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 is not valid.

Solution:-

Since either coin can turn up Head (H) or Tail (T), these are the possible outcomes.

Here, the coin is tossed twice, then the 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 favourable to the event/ Total number of possible outcomes

P(A) = n(A)/n(S)

= ¾

Solution:-

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 favourable to the event/ Total number 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 greater than or equal to 3.

B = {3, 4, 5, 6}

Then, n(B) = 4

P(Event) = Number of outcomes favourable to the event/ Total number 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 than or equal to 1.

C = {1}

Then, n (C) = 1

P(Event) = Number of outcomes favourable to the event/ Total number 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 than 6; then,

D = {0)}

Then, n (D) = 0

P(Event) = Number of outcomes favourable to the event/ Total number 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 than 6; then,

E= (1, 2, 3, 4, 5)

Then, n (E) = 5

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(E) = n(E)/n(S)

= 5/6

Solution:-

From the question, it is given that there are 52 cards in the 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 of spades.

A= 1

Then, n (A) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(A) = n(A)/n(S)

= 1/52

(c) Let us assume ‘B’ be the event of drawing an ace. There are four aces.

Then, n (B)= 4

P(Event) = Number of outcomes favourable to the event/ Total number 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 favourable to the event/ Total number of possible outcomes

∴P(C) = n(C)/n(S)

= 26/52

= ½

Solution:-

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 the sum of numbers as 3.

P = {(1, 2)},

Then, n (P) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(P) = n(P)/n(S)

= 1/12

(ii) Let us assume ‘Q’ be the event having the sum of the number as 12.

Then Q = {(6, 6)}, n (Q) = 1

P(Event) = Number of outcomes favourable to the event/ Total number of possible outcomes

∴P(Q) = n(Q)/n(S)

= 1/12

Solution:-

From the question, it is given that there are four men and six women on the city council.

Here, the total members in the council = 4 + 6 = 10,

Hence, the sample space has 10 points.

∴ n (S) = 10

The number of women is 6 … [given]

Let us assume ‘A’ be the event of selecting a woman.

Then n (A) = 6

P(Event) = Number of outcomes favourable to the event/Total number of possible outcomes

∴P(A) = n(A)/n(S)

= 6/10 … [Divide both numerator and denominators by 2]

= 3/5

Solution:-

Since either coin can turn up Head (H) or Tail (T), these are the possible outcomes.

But, now coin is tossed four times, so the possible sample space 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 win or lose money depending upon 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 and denominator by 4]

= ¼

P (winning ₹ 1) = 6/16 … [Divide both numerator and denominator by 2]

= 3/8

P (winning ₹ 3.50) = 4/16 … [Divide both numerator and denominator by 4]

= ¼

P (winning ₹ 6) = 1/16

= 3/8

Solution:-

Since either coin can turn up Head (H) or Tail (T), these are the possible outcomes.

But, now three coins are tossed, so the possible sample space contains

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 heads.

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

Solution:-

From the question, it is given that 2/11 is the probability of an event A.

i.e., P (A) = 2/11

Then,

P (not A) = 1 – P (A)

= 1 – (2/11)

= (11 – 2)/11

= 9/11

Solution:-

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 the consonant.

n(B)= 7

∴P(B) = n(B)/n(S)

= 7/13