Chapter 36.1 - Examples in Theoretical Probability

In the previous section we saw some examples and learned some new terms in Probability. In this section, we will see a few more examples.

Example 4
One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace  (ii) not be an ace
Solution:
We have already seen the probabilities related to the deck of cards in an earlier chapter. There we saw the various features of a deck of cards. (Details here)
For our present problem, we can straight away begin to write the steps:
Part (i):

• Let P(A) be the probability of 'getting an ace'
Step 1: Write the outcomes

There are 52 possible outcomes. So the denominator is 52
Step 2: Analyse each outcome
• The requirement given is: The card should be an 'ace'.
• There are four aces. So we can write 'favourable' towards four outcomes. 
  ♦ If we get any one of those four, we have an event. So the numerator is 4
Thus P(A) = ⁴⁄₅₂ = ¹⁄₁₃
Part (ii):

• Let P(B) be the probability of 'not getting an ace' 
Step 1: Write the outcomes

There are 52 possible outcomes. So the denominator is 52

Step 2: Analyse each outcome

• The requirement given is: The card should not be an 'ace'.
• There are four aces. So we can write 'favourable' towards (52-4) = 48 outcomes. 
  ♦ If we get any one of those 48, we have an event. So the numerator is 48
Thus P(B) = ⁴⁸⁄₅₂ = ¹²⁄₁₃.
Easy method for part (ii):
1. Event A is the complementary event of event B
• That is., P(A) = P(complement of B)
2. But complementary event of B is (B) 
• So we get: P(A) = P(B)
3. In part (ii), we are trying to find P(B)
• We know that P(B) = 1- P(B)
4. So using the result in step (2) above, we can write:
P(B) = 1 - P(A)
• But in part (i) we obtained P(A) as ¹⁄₁₃
• Thus we get P(B) = (1 - ¹⁄₁₃) = (¹²⁄₁₃)

Example 5
Two players Sangeeta and Reshma play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Solution:
1. Let 'the probability of Sangeeta winning the match' be denoted as P(S)
• So we can write: P(S) = 0.62
2. Let 'the probability of Reshma winning the match' be denoted as P(R)
• We want to find this P(S)
3. The 'event of Reshma winning the match' is same as the 'event of Sangeeta not winning the match'
• In other words, R and S are complimentary 
• That is., R is same as (not S)
    ♦ 'not S' is denoted as 'S'
4. So we can write: R is same as S.
• So P(R) = P(S)
5. We can easily find P(S) because, P(S) is given
• We get: P(S) = (1-P(S)) = (1-0.62) = 0.38
• So from (4), we get: P(R) = P(S) = 0.38

Example 6
Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution:
Part (i):
1. Let Savita's birthday be 'k'
• 'k' can be any value like JAN 12, MAR 25, SEP 14 etc.,
• What ever be the value, it is unique. Because, each day among the 365 days in an year is a unique day
• We want to know the probability of Hamida's birthday to be 'not on k'
2. Let us write the outcomes:
Outcome 1: Hamida's birthday is on JAN 1
Outcome 2: Hamida's birthday is on JAN 2
Outcome 3: Hamida's birthday is on JAN 3
_     _     _     _     _     _     _     _
_     _     _     _     _     _     _     _

Outcome 365: Hamida's birthday is on DEC 31
• No outcomes other than the above 365 can possibly occur
• We say that: The number of all possible outcomes is 365
• So in the denominator, we will have 365
3. We want to present the probability of Hamida's birthday to be 'not on k'
• We will denote it as P(Not same)
• If Hamida's birthday is 'not on k', we have an event
4. Towards all the 365 outcomes in step(2), we will be writing 'favourable' except one.
• We will be writing 'not favourable' towards the outcome in which 'Hamida's birthday is on 'k''
5. So we will be having 364 favourable outcomes.
• Thus, in the numerator, we will be having 364
• So P(Not same) = ³⁶⁴⁄₃₆₅ 
Part (ii):
• For part (ii) we can use the easy method
1. We want to present the probability of Hamida's birthday to be 'on k'
• We will denote it as P(Same)
2. If Hamida's birthday is on 'k',
• It will not be 'Not same'
• That is., if 'Not same' occur, 'Same' will not occur 
    ♦ The condition of 'Same' not occurring is denoted as 'Same'
• Thus we can write: 'Not same' occurring is equivalent to 'Same' occurring.
• So P(Not same) = P(Same)
3. We want P(Same). For that we can use the general relation: P(E) = 1 - P(E)
• So P(Same) = 1 - P(Same)
• Substituting for P(Same) from step (2), we get:
P(Same) = 1 - P(Not same)
4. From part(i), we have P(Not same) = ³⁶⁴⁄₃₆₅
• Thus we get: P(Same) = (1 - ³⁶⁴⁄₃₆₅)= ¹⁄₃₆₅

Example 7
There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
Solution:
Part (i):
1. Let the girls be denoted as: G1, G2, G3, . . . G25
Let the boys be denoted as: B1, B2, B3, . . . B15
2. Let us write the outcomes:
Outcome 1: The drawn card is G1
Outcome 2: The drawn card is G2
Outcome 3: The drawn card is G3
_     _     _     _     _     _     _     _
_     _     _     _     _     _     _     _

Outcome 25: The drawn card is G25

Outcome 26: The drawn card is B1

Outcome 27: The drawn card is B2

Outcome 28: The drawn card is B3

_     _     _     _     _     _     _     _

_     _     _     _     _     _     _     _

Outcome 40: The drawn card is B15

• No outcomes other than the above 40 can possibly occur

• We say that: The number of all possible outcomes is 40

• So in the denominator, we will have 40

3. We want to present the probability of 'a girl's name to be drawn'

• We will denote it as P(G)

• If a girl's name is drawn, we have an event

4. Towards the first 25 outcomes in step(2), we will be writing 'favourable'.

• We will be writing 'not favourable' towards the next 15 outcomes

5. So we will be having 25 favourable outcomes.

• Thus, in the numerator, we will be having 25

• So P(G) = ²⁵⁄₄₀ = ⁵⁄₈ 

---

• This completes part(i) of the problem. Note that, once we understand the basics, we need not write the detailed steps. The following steps will be sufficient:

• Total no. of outcomes = (25+15) = 40

• No. of favourable outcomes = 25

• So P(G) = ^{Number of outcomes favourable to G}⁄_{Number of all outcomes} = ²⁵⁄₄₀ = ⁵⁄₈ 

---

Part (ii):

1. Let the girls be denoted as: G1, G2, G3, . . . G25
Let the boys be denoted as: B1, B2, B3, . . . B15
2. Let us write the outcomes:
Outcome 1: The drawn card is G1
Outcome 2: The drawn card is G2
Outcome 3: The drawn card is G3
_     _     _     _     _     _     _     _
_     _     _     _     _     _     _     _

Outcome 25: The drawn card is G25
Outcome 26: The drawn card is B1

Outcome 27: The drawn card is B2

Outcome 28: The drawn card is B3

_     _     _     _     _     _     _     _

_     _     _     _     _     _     _     _

Outcome 40: The drawn card is B15

• No outcomes other than the above 40 can possibly occur

• We say that: The number of all possible outcomes is 40

• So in the denominator, we will have 40

3. We want to present the probability of 'a boy's name to be drawn'

• We will denote it as P(B)

• If a boy's name is drawn, we have an event

4. Towards the first 25 outcomes in step(2), we will be writing 'unfavourable'.

• We will be writing 'favourable' towards the next 15 outcomes

5. So we will be having 15 favourable outcomes.

• Thus, in the numerator, we will be having 15

• So P(G) = ¹⁵⁄₄₀ = ³⁄₈

---

• This completes part(ii) of the problem. Note that, once we understand the basics, we need not write the detailed steps. The following steps will be sufficient:

• Total no. of outcomes = (25+15) = 40

• No. of favourable outcomes = 15

• So P(B) = ^{Number of outcomes favourable to B}⁄_{Number of all outcomes} = ¹⁵⁄₄₀ = ³⁄₈

Easy method for Part (ii):

• For part (ii) we can use the easy method

1. We want to present the probability of 'a boy's name to be drawn'

• We will denote it as P(B)

2. If 'G' occur, 'B' will not occur 

    ♦ The condition of 'B' not occurring is denoted as 'B'

• Thus we can write: 'G' occurring is equivalent to 'B' occuring.

• So P(G) = P(B)

3. We want P(B). For that we can use the general relation: P(E) = 1 - P(E)

• So P(B) = 1 - P(B)

• Substituting for P(B) from step (2), we get:

P(B) = 1 - P(G)

4. From part(i), we have P(G) = ⁵⁄₈

• Thus we get: P(B) = (1 - ⁵⁄₈) = ³⁄₈

■ Note that, once we understand the basics, this easy method can also be written in a lesser number of steps:

1. If 'G' occur, 'B' will not occur 

    ♦ The condition of 'B' not occurring is denoted as 'B'

• That is., 'G' occurring is same as 'B' occurring

• So P(G) = P(B)

2. We want P(B), which is equal to [1-P(B)]  

• But from step (1), we have P(B) = P(G)

• So P(B) = [1-P(B)] = [1-P(G)] = [1-⁵⁄₈] = ³⁄₈

Example 8:

A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) white? (ii) blue? (iii) red?

Solution:

Part (i):

1. Let the 3 blue marbles be denoted as: B1, B2 and B3

• Let the 2 white marbles be denoted as: W1 and W2

• Let the 4 red marbles be denoted as: R1, R2, R3 and R4

2. Let us write the outcomes:

Outcome 1: The drawn marble is B1

Outcome 2: The drawn marble is B2

Outcome 3: The drawn marble is B3

Outcome 4: The drawn marble is W1

Outcome 5: The drawn marble is W2

Outcome 6: The drawn marble is R1

Outcome 7: The drawn marble is R2

Outcome 8: The drawn marble is R3

Outcome 9: The drawn marble is R4

• No outcomes other than the above 9 can possibly occur

• We say that: The number of all possible outcomes is 9

• So in the denominator, we will have 9

3. We want to present the probability of 'a white marble to be drawn'

• We will denote it as P(W)

• If a white marble is drawn, we have an event

4. Towards the first 3 outcomes in step(2), we will be writing 'not favourable'.
• Towards the next 2 outcomes in step(2), we will be writing 'favourable'.
• Towards the last 4 outcomes in step(2), we will be writing 'not favourable'.

5. So we will be having 2 favourable outcomes.

• Thus, in the numerator, we will be having 2

• So P(W) = ²⁄₉

■ The above steps can be shortened as:

• Total no. of outcomes = (3+2+4) = 9

• No. of favourable outcomes = 2

• So P(W) = ^{Number of outcomes favourable to W}⁄_{Number of all outcomes} = ²⁄₉

Part (ii):

• Total no. of outcomes = (3+2+4) = 9

• No. of favourable outcomes = 3

• So P(B) = ^{Number of outcomes favourable to B}⁄_{Number of all outcomes} = ³⁄₉

Part (iii):

• Total no. of outcomes = (3+2+4) = 9

• No. of favourable outcomes = 4

• So P(R) = ^{Number of outcomes favourable to R}⁄_{Number of all outcomes} = ⁴⁄₉.

■ Note that P(W) + P(B) + P(R) = [²⁄₉ + ³⁄₉ + ⁴⁄₉]  = [⁹⁄₉] = 1

Example 9

Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that she gets at least one head?

Solution:

1. Let us denote the sides:

• Head of Re 1 is 1H • Tail of Re 1 is 1T

• Head of Re 2 is 2H •Tail of Re 2 is 2T

2. Let us write the outcomes:

• Outcome 1: Re 1 lands with H on upper face and Re 2 also lands with H on upper face

    ♦ That is., we have: [1H, 2H] 

• Outcome 2: Re 1 lands with T on upper face and Re 2 also lands with T on upper face

    ♦ That is., we have: [1T, 2T] 

• Outcome 3: Re 1 lands with H on upper face and Re 2 lands with T on upper face

    ♦ That is., we have: [1H, 2T] 

• Outcome 4: Re 1 lands with T on upper face and Re 2 lands with H on upper face

    ♦ That is., we have: [1T, 2H]

• No outcomes other than the above 4 can possibly occur

• We say that: The number of all possible outcomes is 4

• So in the denominator, we will have 4

3. We want to present the probability of 'at least one head'

• We will denote it as P(H)

• If at least one head is obtained, we have an event 

4. Towards the first outcome in step(2), we will be writing 'favourable'.

    ♦ This is because there is at least one head

• Towards the second outcome in step(2), we will be writing 'not favourable'.

    ♦ This is because there is not even a single head

• Towards the third outcome in step(2), we will be writing 'favourable'.

    ♦ This is because there is at least one head

• Towards the fourth outcome in step(2), we will be writing 'favourable'.

    ♦ This is because there is at least one head

5. So we will be having 3 favourable outcomes.

• Thus, in the numerator, we will be having 3

• So P(H) = ³⁄₄ 

■ Can we apply P(H) = [1-P(H)] in this problem?

Let us try:

• (H) indicates an event of 'no head'

• Out of the 4 possible outcomes, only one (outcome 3) is favourable for 'no head'

• So P(H) = ¹⁄₄

• Thus we get: P(H) = [1-P(H)] = [1 - ¹⁄₄] = ³⁄₄

---

In this section, we saw 3 examples and learned some new terms related to Probability. In the next section, we will see more examples.

PREVIOUS      CONTENTS       NEXT

 
                        Copyright©2018 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 28.2- Tossing two coins simultaneously

In the previous section we saw the results when the number of times a die is rolled is increased. In this section we will see such experiments related to 'tossing two coins simultaneously'.

Let us do an experiment. We will call it experiment IIIA:
1. Take two coins. 
(i) Toss them simultaneously. The reading may be any one of the following:
HH, TT and HT
Note that, in the result HT, it does not matter which coin gives H and which one T
(ii) What ever be the reading, note that reading on a piece of paper. 
• (i) and (ii) constitutes one cycle of our experiment.
2. Repeat the cycle 10 times. 
3. In the note book, tabulate the readings as shown in table 28.5 below 
Table 28.5

Number of times the two coins are tossed	Number of times no heads comes up (TT)	Number of times one head comes up (HT)	Number of times two heads come up (HH)
10	2	6	1

4. Determine the following ratios:
• ^{Number of times no head turned up}⁄_{Total number of times the two coins are tossed}
• ^{Number of times one head turned up}⁄_{Total number of times the two coins are tossed}
• ^{Number of times two heads turned up}⁄_{Total number of times the two coins are tossed}
• In our present case:
    ♦ the 1^st ratio is ²⁄₁₀ = 0.2
    ♦ the 2^nd ratio is ⁶⁄₁₀ = 0.6
    ♦ the 3^rd ratio is ¹⁄₁₀ = 0.1
5. Once the ratios in (4) are determined, the experiment IIIA is complete

---

But our work is not over

■ Repeat the above experiment. We will call it experiment IIIB.
• For this experiment IIIB, the number of cycles in (2) must be 20
• Let the readings be as shown below:

Number of times the two coins are tossed	Number of times no heads comes up (TT)	Number of times one head comes up (HT)	Number of times two heads come up (HH)
20	5	11	4

• In this case:
    ♦ the 1^st ratio is ⁵⁄₂₀ = 0.25
    ♦ the 2^nd ratio is ¹¹⁄₂₀ = 0.55
    ♦ the 3^rd ratio is ⁴⁄₂₀ = 0.2
• When the ratios in (4) are determined, the experiment IIIB is over
■ Once again repeat the experiment. We will call it experiment IIIC
• For this experiment IIIC, the number of cycles in (2) must be 30
• Let the readings be as shown below:

Number of times the two coins are tossed	Number of times no heads comes up (TT)	Number of times one head comes up (HT)	Number of times two heads come up (HH)
30	8	17	5

• In this case:
    ♦ the 1^st ratio is ⁸⁄₃₀ = 0.267
    ♦ the 2^nd ratio is ¹⁷⁄₃₀ = 0.567
    ♦ the 3^rd ratio is ⁵⁄₃₀ = 0.167
• When the ratios in (4) are determined, the experiment IIIC is over

---

So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the six ratios:

1. We know that, the probability of obtaining TT, HT and HH when tossing two coins simultaneously are 0.25, 0.5 and 0.25 respectively

2. But these are theoretical values. If they are always obtained in the real life also, we will get results such as these:

• Toss two coins simultaneously 12 times
    ♦ TT will be obtained 3 times
    ♦ HT will be obtained 6 times
    ♦ HH will be obtained 3 times
• Toss two coins simultaneously 16 times
    ♦ TT will be obtained 4 times
    ♦ HT will be obtained 8 times
    ♦ HH will be obtained 4 times

3. But we never get such exact values.
4. However, as the number of trials increase, each of the 3 ratios become closer and closer to their respective values 0.25, 0.5 and 0.25

5. If, instead of 10,20 or 30 times, if we toss it for a 'very large number of times, n', then:
• ^{Number of times TT is obtained}⁄_n = 0.25
• ^{Number of times HT is obtained}⁄_n = 0.5
• ^{Number of times HH is obtained}⁄_n = 0.25

We are trying to prove this using our present experiments

---

• From the three experiments, we have three sets of ratios. One set from each experiment

• Each set has six ratios:
    ♦ ^{Number of times TT is obtained}⁄_n
    ♦ ^{Number of times HT is obtained}⁄_n
    ♦ ^{Number of times HH is obtained}⁄_n
■ Let us now analyse the ratios:
1^st ratio when number of trials is 10 = 0.2
1^st ratio when number of trials is 20 = 0.25
1^st ratio when number of trials is 30 = 0.267
■ As the number of trial increases, the 1^st ratio gets closer and closer to 0.25
Consider the 2^nd ratio:
2^nd ratio when number of trials is 10 = 0.6
2^nd ratio when number of trials is 20 = 0.55
2^nd ratio when number of trials is 30 = 0.567
■ As the number of trial increases, the 2^nd ratio gets closer and closer to 0.5
Consider the 3^rd ratio:
3^rd ratio when number of trials is 10 = 0.1
3^rd ratio when number of trials is 20 = 0.2
3^rd ratio when number of trials is 30 = 0.167
■ As the number of trial increases, the 3^rd ratio gets closer and closer to 0.25

---

• We can increase the 'number of trials' to a 'considerably large value'  by using groups as we saw in the previous sections. Students may try it themselves.
• It will become clear that the ratios get closer and closer to 0.25, 0.5 and 0.25

---

In the next section, we will see how the above findings can be used to solve practical problems.  

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Maths lessons. blogspot.in - All Rights Reserved