In the previous section,
we saw how to calculate the Probability of the event "not A".
We saw two solved examples also. In this section, we will see two more solved examples.
Solved example 16.12
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.
(b) Atleast one of them will not qualify the examination and
(c) Only one of them will qualify the examination.
Solution:
First we will write two basic points. We will number them as πΌ and Ξ²:
Point πΌ:
This can be written in 6 steps:
1. Consider the Venn diagram shown in fig.16.2(a) below. It is already familiar to us.
 |
| Fig.6.2 |
2. (Red ∪ Blue) gives set A.
In our present case, A is the event: Anil qualifies in the examination.
3. (Green ∪ Blue) gives set B.
In our present case, B is the event: Ashima qualifies in the examination.
4. Blue is the set (A∩B).
In our present case, (A∩B) is the event: Both Anil and Ashima qualify in the examination.
5. (Red ∪ Blue ∪ Green) gives the set (A∪B).
In our present case, (A∪B) is the event: Anil or Ashima qualifies in the examination.
6. Consider the portion outside (A∪B), but inside the rectangle. That portion gives the set (A∪B)’. It is the portion shown in yellow color in fig.16.2(b) above.
In our present case, (A∪B)’ is the event: Both Anil and Ashima does not qualify in the examination.
Point Ξ²:
This can be written in 6 steps:
1. Consider the rectangle S. Imagine that, there are a large number of elements dispersed inside the rectangle. Each of those elements is an outcome.
2. Consider all the outcomes in S. Some of those outcomes are present within (Red ∪ Blue).
•
If the outcome of the experiment is from this region, we say that:
Anil has qualified.
•
The probability for the outcome to be from this region is given in the question. P(A) = 0.05
3. Consider all the outcomes in S. Some of those outcomes are present within (Green ∪ Blue).
•
If the outcome of the experiment is from this region, we say that:
Ashima has qualified.
•
The probability for the outcome to be from this region is given in the question. P(B) = 0.1
4. Consider all the outcomes in S. Some of those outcomes are present within Blue.
•
If the outcome of the experiment is from this region, we say that:
Both Anil and Ashima have qualified.
•
The probability for the outcome to be from this region is given in the question. P(A∩B) = 0.02
5. A and B are not disjoint sets. This can be proved in 3 steps:
(i) If A and B are disjoint sets, (A∩B) = Ξ¦
(ii) Probability of Ξ¦ is zero.
(iii) But according to the question, P(A∩B) is not zero. It is 0.02
6. We said that, a large number of outcomes are dispersed within the rectangle S.
• In our present case,
♦ we do not know how many such outcomes are there.
♦ we do not know the probabilities of each of those outcomes.
• If we knew the number of outcomes and their probabilities, we could calculate P(A), P(B) etc.,
• But as the reader may have already noted, in our present case, we do not need them because, P(A), P(B) and P(A∩B) are already given.
Now we can answer the questions.
Part (i): Both Anil and Ashima will not qualify.
1. Consider the region (A∪B). It is made up of three regions: red, blue and green.
• If the outcome is from the red region, Anil qualifies. So “Both Anil and Ashima will not qualify” is not satisfied.
• If the outcome is from the blue region, Anil and Ashima qualifies. So “Both Anil and Ashima will not qualify” is not satisfied.
• If the outcome is from the green region, Ashima qualifies. So “Both Anil and Ashima will not qualify” is not satisfied.
• It is clear that, we must discard (A∪B).
2. Consider the region outside (A∪B), but inside the rectangle.
•
We know that, such a region is the compliment of set (A∪B). We denote it as (A∪B)’. It is the yellow region of the Venn diagram in fig.16.2(b) above.
• If the outcome is from (A∪B)', we say that:
Both Anil and Ashima will not qualify.
3. The probability for the outcome to be from (A∪B)' is: P(A∪B)'
•
So our aim is to find P(A∪B)'
4. It is clear that, (A∪B) and (A∪B)’ are mutually exclusive and exhaustive events.
•
We can write:
(A∪B)∪(A∪B)' = S.
•
Based on this, we can write the calculations as follows:
$\begin{array}{ll}
{}&{\rm{(A \cup B) \cup (A \cup B)'}}
& {~=~}& {\rm{S}}
&{} \\
{\Rightarrow}&{\rm{P \Bigl((A \cup B) \cup (A \cup B)' \Bigr)}}
& {~=~}& {\rm{P(S)}}
&{} \\
{\Rightarrow}&{\rm{P(A \cup B) + P(A \cup B)'}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - I}}}}
&{} \\
{\Rightarrow}&{\rm{P(A) + P(B) - P(A \cap B) + P(A \cup B)'}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - II}}}}
&{} \\
{\Rightarrow}&{\rm{0.05 + 0.1 - 0.02 + P(A \cup B)'}}
& {~=~}& {\rm{1}}
&{} \\
{\Rightarrow}&{\rm{0.13 + P(A \cup B)'}}
& {~=~}& {\rm{1}}
&{} \\
{\Rightarrow}&{\rm{P(A \cup B)'}}
& {~=~}& {\rm{1 - 0.13}}
&{} \\
{\Rightarrow}&{\rm{P(A \cup B)'}}
& {~=~}& {\rm{0.87}}
&{} \\
\end{array}$
◼ Remarks:
•
Line marked as I:
In this line we use the formula:
P(E∪F) = P(E) + P(F)
Where E and F are disjoint sets.
•
Line marked as II:
In this line we use the formula:
P(E∪F) = P(E) + P(F) - P(E∩F)
Where E and F are not disjoint sets.
Part (ii): Atleast one of them will not qualify the examination
1. Consider the blue region.
If the outcome is from this region, it means that both Anil and Ashima qualifies for the examination. So we have to discard this region.
2. Consider the region outside blue but inside the rectangle. This region is (A∩B)’
•
This (A∩B)’ is made up of three regions:
(i) Red region (ii) Green region (iii) yellow region.
3. Let us examine each of the three regions.
(i) If the outcome is from the red region, then Anil qualifies but Ashima does not qualify. So “atleast one of them will not qualify” is satisfied.
(ii) If the outcome is from the green region, then Ashima qualifies but Anil does not qualify. So “atleast one of them will not qualify” is satisfied.
(iii) If the outcome is from yellow region, then both do not qualify. So “atleast one of them will not qualify” is satisfied.
4. So (A∩B)’ is our required region.
•
The probability for the outcome to be from this region is: P(A∩B)’
5. We have:
$\begin{array}{ll}
{}&{\rm{P(A \cap B)'}}
& {~=~}& {\rm{1 - P(A \cap B)}} &{} \\
{}&{}
& {~=~}& {\rm{1 - 0.02}} &{} \\
{}&{}
& {~=~}& {\rm{0.98}} &{} \\
\end{array}$
Part (iii): Only one of them will qualify the examination.
1. Let us examine each region in fig.16.2 above.
(i) The red region.
If the outcome is from red, then only Anil qualifies. So this region can be considered for our answer.
(ii) The blue region.
If the outcome is from blue, then both qualify. So this region cannot be considered for our answer.
(iii) Green region.
If the outcome is from green, then only Ashima qualifies. So this region can be considered for our answer.
(iv) Yellow region.
If the outcome is from yellow , then neither Anil nor Ashima qualifies. So this region cannot be considered for our answer.
2. Based on the above step, we can write:
The only regions than can be considered are: red and green.
3. We can create a new set: (red ∪ green)
If the outcome is from this union, there are two possibilities:
(i) outcome is from red.
Then only Anil qualifies. So “only one of them will qualify” is satisfied.
(ii) outcome is from green.
Then only Ashima qualifies. So “only one of them will qualify” is satisfied.
4. So (red ∪ green) is our required region.
♦ Red is (A-B)
♦ Green is (B-A)
• So (A-B)∪(B-A) is our required region.
5. Probability for the outcome to be from this region is: $P \Bigl((A-B) \cup (B-A) \Bigr)$
• (A-B) and (B-A) are disjoint sets. So we can write:
$P \Bigl((A-B) \cup (B-A) \Bigr) = P \Bigl((A-B)\Bigr) + P \Bigl((B-A)\Bigr)$
• So we have to calculate $P \Bigl((A-B)\Bigr) ~\text{and}~ P \Bigl((B-A)\Bigr)$
6. First we will calculate $P \Bigl((A-B)\Bigr)$
$\begin{array}{ll}
{}&{\rm{A}}
& {~=~}& {\rm{(A-B) \cup (A \cap B)~\color{green}{\text{- - - I}}}}
&{} \\
{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{P \Bigl((A-B) \cup (A \cap B)\Bigr)}}
&{} \\
{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{P \Bigl((A-B)\Bigr) ~+~ P \Bigl((A \cap B)\Bigr)~\color{green}{\text{- - - II}}}}
&{} \\
{\Rightarrow}&{\rm{0.05}}
& {~=~}& {\rm{P \Bigl((A-B)\Bigr) ~+~ 0.02}}
&{} \\
{\Rightarrow}&{\rm{P \Bigl((A-B)\Bigr)}}
& {~=~}& {\rm{0.05~-~ 0.02}}
&{} \\
{\Rightarrow}&{\rm{P \Bigl((A-B)\Bigr)}}
& {~=~}& {\rm{0.03}}
&{} \\
\end{array}$
◼ Remarks:
• Line marked as I:
Set A is the union of red and blue.
• Line marked as II:
We are able to simply add the individual probabilities because, (A-B) and (A∩B) are disjoint sets.
7. Next we will calculate $P \Bigl((B-A)\Bigr)$
$\begin{array}{ll}
{}&{\rm{B}}
& {~=~}& {\rm{(B-A) \cup (A \cap B)~\color{green}{\text{- - - I}}}}
&{} \\
{\Rightarrow}&{\rm{P(B)}}
& {~=~}& {\rm{P \Bigl((B-A) \cup (A \cap B)\Bigr)}}
&{} \\
{\Rightarrow}&{\rm{P(B)}}
& {~=~}& {\rm{P \Bigl((B-A)\Bigr) ~+~ P \Bigl((A \cap B)\Bigr)~\color{green}{\text{- - - II}}}}
&{} \\
{\Rightarrow}&{\rm{0.1}}
& {~=~}& {\rm{P \Bigl((B-A)\Bigr) ~+~ 0.02}}
&{} \\
{\Rightarrow}&{\rm{P \Bigl((B-A)\Bigr)}}
& {~=~}& {\rm{0.1~-~ 0.02}}
&{} \\
{\Rightarrow}&{\rm{P \Bigl((B-A)\Bigr)}}
& {~=~}& {\rm{0.08}}
&{} \\
\end{array}$
◼ Remarks:
• Line marked as I:
Set B is the union of green and blue.
• Line marked as II:
We are able to simply add the individual probabilities because, (B-A) and (A∩B) are disjoint sets.
8. Substituting the results from (6) and (7) in (5), we get:
$P \Bigl((A-B) \cup (B-A) \Bigr) = 0.03 + 0.08 = 0.11$
Solved example 16.13
A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man? (b) one man? (c) two men?
Solution:
◼ There are two men (M1 & M2) and two women (W1 & W2)
♦ So there is a total of four persons
◼ From that four, two persons can be selected in $\rm{{}^4 C_2}$ ways.
♦ So the number of possible outcomes = $\rm{{}^4 C_2}$
♦ We can write: n(S) = $\rm{{}^4 C_2}$
♦ All the $\rm{{}^4 C_2}$ outcomes are equally likely.
♦ Some of those outcomes are: (M1, W1), (W2, M1), etc.,
• Now we can do the calculations:
Part (i):
1. Let A be the event: Getting an outcome with no man.
2. Since there is to be no man, we must not consider M1 and M2 while making the selections.
• That means, we must consider W1 and W2 only.
• Two women can be selected from two women in $\rm{{}^2 C_2}$ ways.
• So n(A) = $\rm{{}^2 C_2}$
3. Since all outcomes are equally likely, we get:
$\rm{P(A) = \frac{n(A)}{n(S)} = \frac{{}^2 C_2}{{}^4 C_2} = \frac{1}{6}}$
Part (ii):
1. Let B be the event: Getting an outcome with one man.
2. Since there is to be exactly one man, the other person in the committee will be woman.
• One man can be selected from two men in $\rm{{}^2 C_1}$ ways.
• One woman can be selected from two women in $\rm{{}^2 C_1}$ ways.
• Together, they can be selected in $\rm{{}^2 C_1 \times {}^2 C_1}$ ways.
• So n(B) = $\rm{{}^2 C_1 \times {}^2 C_1}$
3. Since all outcomes are equally likely, we get:
$\rm{P(B) = \frac{n(B)}{n(S)} = \frac{{}^2 C_1 \times {}^2 C_1}{{}^4 C_2} = \frac{2 \times 2}{6} = \frac{2}{3}}$
Part (iii):
1. Let C be the event: Getting an outcome with two men.
2. Since there is to be two men, we must not consider W1 and W2 while making the selections.
• That means, we must consider M1 and M2 only.
• Two men can be selected from two men in $\rm{{}^2 C_2}$ ways.
• So n(C) = $\rm{{}^2 C_2}$
3. Since all outcomes are equally likely, we get:
$\rm{P(C) = \frac{n(C)}{n(S)} = \frac{{}^2 C_2}{{}^4 C_2} = \frac{1}{6}}$
Link to a few more solved examples is given below:
In the next section, we will see some miscellaneous examples.
