Chapter 14: Probability

Ans: 

(a) Probability should be less than 1 and positive.

\[P\left( E \right) = 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6\]

\[P\left( E \right) = 1\]

Therefore, the given assignment is valid.

b) Probability should be less than 1 and positive.

\[P\left( E \right) = \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7}\]

\[P\left( E \right) = 1\]

Therefore, the given assignment is valid.

c) Probability should be less than 1 and positive.

\[P\left( E \right) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7\]

\[P\left( E \right) = 2.8 > 1\]

Here probability is greater than 1.

Therefore, the given assignment is not valid.

d) Probability can never be negative.

Therefore, the assignment is not valid.

e) Probability should be less than 1 and positive.

\[P\left( E \right) = \dfrac{1}{{14}} + \dfrac{2}{{14}} + \dfrac{3}{{14}} + \dfrac{4}{{14}} + \dfrac{5}{{14}} + \dfrac{6}{{14}} + \dfrac{{15}}{{14}}\]

\[P\left( E \right) = \dfrac{{28}}{{14}} > 1\]

Therefore, the assignment is not valid.

Ans: Here coin is tossed twice, then sample space is \[S = \left( {TT,HH,TH,HT} \right)\]

Number of possible outcomes \[n\left( S \right) = 4\]

Let A be the event of getting at least one tail

\[\therefore n\left( A \right) = 3\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{3}{4}\]

Ans: Here die is thrown, therefore the sample space of outcomes will be \[S = \left\{ {1,2,3,4,5,6} \right\}\]

Let A be the event of getting a prime number,

\[A = \left\{ {2,3,5} \right\}\]

Then, \[n\left( A \right) = 3\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{3}{6}\]

\[P\left( A \right) = \dfrac{1}{2}\]

Ans: Let B be the event of getting a number greater than or equal to 3,

\[B = \left\{ {3,4,5,6} \right\}\]

Then, \[n\left( B \right) = 4\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( B \right) = \dfrac{4}{6}\]

\[P\left( B \right) = \dfrac{2}{3}\]

Ans: Let A be the event of getting a number less than or equal to 1,

\[A = \left\{ 1 \right\}\]

Then, \[n\left( A \right) = 1\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{6}\]

Ans: Let A be the event of getting a number more than 6, then

\[A = \left\{ 0 \right\}\]

Then, \[n\left( A \right) = 0\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{0}{6}\]

\[P\left( A \right) = 0\]

Ans: Let E be the event of getting a number less than 6, then

\[A = \left( {1,2,3,4,5} \right)\]

Then, \[n\left( A \right) = 5\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{5}{6}\]

Ans: Number of points in the sample space = 52

\[\therefore {\text{ }}n\left( S \right) = 52\]

Ans: Let A be the event of drawing an ace of spades.

\[A = \left\{ 1 \right\}\]

Then, \[n\left( A \right) = 1\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{{52}}\]

(c) Calculate the probability that the card is 

Ans: Let A be the event of drawing an ace. There are four aces.

Then, \[n\left( A \right) = 4\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{4}{{52}}\]

\[P\left( A \right) = \dfrac{1}{{13}}\]

Ans: Let C be the event of drawing a black card. There are 26 black cards.

Then, \[n\left( A \right) = 26\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{{26}}{{52}}\]

\[P\left( A \right) = \dfrac{1}{2}\]

Ans: As die and a coin marked 1 is tossed therefore, the sample space of possible outcomes will be,

\[S = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)} \right\}\]

Therefore, \[n\left( S \right) = 12\]

Let A be the event having sum of numbers as 3.

\[A = \left\{ {\left( {1,2} \right)} \right\}\]

\[n\left( A \right) = 1\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{{12}}\]

Ans: Let A be the event having sum of number as 12.

\[A = \left\{ {\left( {6,6} \right)} \right\}\]

\[n\left( A \right) = 1\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{{12}}\]

Ans: \[{\text{Total members in the council}} = 4 + 6\]

\[{\text{Total members in the council}} = 10\]

Therefore,

 \[n\left( S \right) = 10\]

Number of women are 6 

Let A be the event of selecting a woman

Therefore, \[n\left( A \right) = 6\]

\[P\left( E \right){\text{ }} = {\text{ }}\dfrac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{6}{{10}}\]

\[P\left( A \right) = \dfrac{3}{5}\]

Ans: Here, 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}

According to question, a person will win or lose money depending up on the face of the coin so,

(i) For 4 heads \[ = 1 + 1 + 1 + 1\] = ₹ 4

Therefore, he wins ₹ 4

(ii) For 3 heads and 1 tail \[ = 1 + 1 + 1--1.50\]

Therefore, he wins ₹ 1.50

(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50

= 2 – 3

= – ₹ 1

Therefore, he will be losing ₹ 1

(iv) For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50

= 1 – 4.50

= – ₹ 3.50

Therefore, he will be losing Rs. 3.50

(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50

= – ₹ 6

Thererfore, he will be losing Rs. 6

So, the new sample space will be

\[S = {\text{ }}\left\{ {4,1.50,1.50,1.50,1.50,--1,--1,--1,--1,--1,--1,--3.50,--3.50,--3.50,--3.50,--6} \right\}\]

Then, \[n\left( S \right) = 16\]

P (winning ₹ 4) \[ = \dfrac{1}{{16}}\]

P (winning ₹ 1.50) \[ = \dfrac{4}{{16}}\]

P (winning ₹ 1.50) \[ = \dfrac{1}{4}\]

P (loosing ₹ 1) \[ = \dfrac{6}{{16}}\]

P (loosing ₹ 1) \[ = \dfrac{3}{8}\]

P (loosing ₹ 3.50) \[ = \dfrac{4}{{16}}\]

P (loosing ₹ 3.50) \[ = \dfrac{1}{4}\]

P (loosing ₹ 6) \[ = \dfrac{1}{{16}}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting 3 heads

\[n\left( A \right) = 1\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting 2 heads

\[n\left( A \right) = 3\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{3}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting at least 2 head

\[n\left( A \right) = 4\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{4}{8}\]

\[P\left( A \right) = \dfrac{1}{2}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting at most 2 heads

\[n\left( A \right) = 7\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{7}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting no heads

\[n\left( A \right) = 1\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting 3 tails

\[n\left( A \right) = 1\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting exactly 2 tails

\[n\left( A \right) = 3\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{3}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting no tails

\[n\left( A \right) = 1\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{8}\]

Ans: Three coin is tossed so the possible sample space contains,

\[S = \left\{ {HHH,HHT,HTH,THH,TTH,HTT,TTT,THT} \right\}\]

\[n\left( S \right) = 8\]

Let A be the event of getting at most 2 tails

\[n\left( A \right) = 7\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{7}{8}\]

Ans: According to question, it is given that, 2/11 is the probability of an event A,

i.e. \[P\left( A \right) = \dfrac{2}{{11}}\]

Therefore,

\[P\left( {{\text{not }}A} \right) = 1 - \dfrac{2}{{11}}\]

\[P\left( {{\text{not }}A} \right) = \dfrac{9}{{11}}\]

Ans: Total letters in the given word is 13

\[n\left( S \right) = 13\]

Number of vowels in the given word is 6

Let A be the event of selecting a vowel

\[n\left( A \right) = 6\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{6}{{13}}\]

Ans: Total letters in the given word is 13

\[n\left( S \right) = 13\]

Number of consonants in the given word = 7

Let A be the event of selecting the consonant

\[n\left( A \right) = 7\]

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{7}{{13}}\]