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Theorem of total probability - class-XI

Description: theorem of total probability
Number of Questions: 18
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Tags: probability - iii probability and probability distribution probability introduction of probability theory probability using tree diagrams and venn diagrams maths
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There are 50 marbles of 3 colors: blue yellow and black The probability of picking up a blue marble is 3/10 and that of picking up a yellow marble is 1/2 The probability of picking up a black ball is 

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. 1/5

  2. 1/10

  3. 1/4

  4. 4/5

Show answer
Correct Option: A
Explanation:

P(blue)=$\displaystyle \frac{3}{10}=\frac{15}{50},$i.e., there are 15 blue marbles
P(yellow)=$\displaystyle \frac{1}{2}=\frac{25}{50},$i.e., there are 25 yellow marbles
$\displaystyle \therefore $ Number of black marbles=10
$\displaystyle \therefore $ P(black)=$\displaystyle \frac{10}{50}=\frac{1}{5}$

Difference between sample space and subset of sample space is considered as 

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. numerical complementary events.

  2. equal compulsory events.

  3. complementary events.

  4. compulsory events.

Show answer
Correct Option: C
Explanation:
The set of all the possible outcomes is called the sample space of the experiment and is usually denoted by S. 
Any subset E of the sample space S
Difference between sample space and subset of sample space is considered as complementary events.

We draw two cards from a deck of shuffled cards without replacement. Find the probability of getting the second card a queen.

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $\dfrac{1}{13}$

  2. $\dfrac{2}{13}$

  3. $\dfrac{5}{13}$

  4. None of these

Show answer
Correct Option: A
Explanation:
There are two cases here:
Case 1: First card chosen is a queen
$\dfrac 4{52}\times \dfrac 3{51}=\dfrac {1}{221}$
Case 2: First card chosen is not a queen.
$\dfrac {48}{52}\times \dfrac 4{51}=\dfrac {16}{221}$
Adding both the cases, we get the probability of getting the second card a queen. $\dfrac {17}{221} =\dfrac  4{52} = \dfrac 1{13}$

Which of the following is true regarding law of total probability?

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. It is a fundamental rule relating marginal probabilities to conditional probabilities.

  2. It expresses the total probability of an outcome which can be realized via several distinct events

  3. Both are correct

  4. None of these

Show answer
Correct Option: C
Explanation:
Law of total probability states that,
        If $B _1,B _2,B _3,...$ is a partition of the sample space $S$, then for any event $A$ we have
$P(A)=\sum _iP(A\cap B _i)=\sum _i P(A|B _i)P(B _i)$

That is, the law (or formula) of total probability is a fundamental rule relating marginal probabilities to conditional probabilities. It expresses the total probability of an outcome which can be realized via several distinct events and hence the name.

Thus option $(C)$ is correct.

I have three bags that each contain $100$ marbles- Bag $1$ has $75$ red and $25$ blue marbles, Bag $2$ has $60$ red and $40$ blue marbles, Bag $3$ has $45$ red and $55$ blue marbles. I choose one of the bags at random and then pick a marble from the chosen bag, also at random. What is the probability that the chosen marble is red?

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $0.60$

  2. $0.40$

  3. $0.50$

  4. None of these

Show answer
Correct Option: A
Explanation:

Given that bag 1 contains $75$ red and $25$ blue marbles

Given that bag 2 contains $60$ red and $40$ blue marbles
Given that bag 3 contains $45$ red abd $55$ blue marbles
Now the probability of choosing red ball from bag 1 is $ \dfrac{1}{3} \times \dfrac{75}{100}$
Now the probability of choosing red ball from bag 2 is $ \dfrac{1}{3} \times \dfrac{60}{100}$
Now the probability of choosing red ball from bag 3 is $ \dfrac{1}{3} \times \dfrac{45}{100}$
Now the probability of choosing red ball is $ \dfrac{1}{3}\left(\dfrac{75+60+45}{100}\right)=0.6$ 
Hence, option $A$ is correct.

For a random experiment, all possible outcomes are called

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. numerical space.

  2. event space.

  3. sample space.

  4. both b and c.

Show answer
Correct Option: D
Explanation:

We know that,

An outcome is a result of a random experiment. 
The set of all possible outcomes is called the sample space.
The subset of the sample space is called event space.
Thus, for a random experiment, all possible outcomes are called sample space and event space.
Thus option $(D)$ is correct.

Tossing a coin is an example of .........

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. Infinite discrete sample space

  2. Finite sample space

  3. Continuous sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

A coin is tossed.

The possible outcomes are "$\text{head,tail}$"
Thus $ S={H,T}$
We get a finite sample space.
Thus tossing a coin is an example of finite sample space.

The term law of total probability is sometimes taken to mean the ____

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. Law of total expectation

  2. Law of alternatives

  3. Law of variance

  4. None of these

Show answer
Correct Option: B
Explanation:

The term law of total probability is sometimes taken to mean the law of alternatives, which is a special case of the law of total probability applying to discrete random variables.

Hence, option B is correct.

The events $E _1, E _2, ........$ represents the partition of the sample space $S$, if they are:

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. pairwise disjoint

  2. exhaustive

  3. have non-zero probabilities

  4. All are correct

Show answer
Correct Option: D
Explanation:

A set of events $E _1 , E _2 ,...$  is said to represent a partition of a sample space $S$, if 


$(a)$  $E _i \cap E _j = \phi, i\neq j; i, j = 1, 2, 3,..., n$  (pairwise disjoint)

$(b)$ $E _i \cup E _2 \cup ... \cup E _n = S$ (exhaustive)

$(c)$ Each $E _i \neq \phi, i.e, P(E _i) > 0$ for all $i = 1, 2, ..., n$ (have non-zero probabilities)

That is the events should be pairwise disjoint, exhaustive and should have non zero probabilities.

Hence, option D is correct.

The experiment is to repeatedly toss a coin until first tail shows up. Identify the type of the sample space.

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. Finite sample space

  2. Continuous sample space

  3. Infinite discrete sample space

  4. None of these

Show answer
Correct Option: C
Explanation:

The experiment is to repeatedly toss a coin until first tail shows up.

A coin is tossed. The possible outcomes are $head-H, tail-T$.
Look into the following table:

 $1^{st}$ toss  $2^{nd}$ toss  $3^{rd}$ toss  $4^{th}$ toss $5^{th}$ toss 
 $T$ $-$ $-$ $-$  $-$
 $H$  $T$  $-$ $-$  $-$
 $H$  $H$  $T$  $-$ $-$
 $H$  $H$  $H$  $T$  $-$
 $H$  $H$  $H$  $H$  $T$

If we get $T$ at first toss, then our experiment ends.
Otherwise second toss. If we get $T$, out experiment ends. If not the process continues till we end up in tail.
Hence the possible outcomes are sequences of $H$ that, if finite, end with a single $T$, and an infinite sequence of $H$.
Therefore, $ S={T,HT,HHT,HHHT,HHHHT,..., {HHHH....}}$
Thus we get a infinite but countable(depends on the number of toss) sample space.
As we shall see elsewhere, this is a remarkable space that contains a not impossible event whose probability is $0$. 
We know that, "a discrete sample space is one with a finite or countably infinite number of possible values."
Hence, it is a infinite discrete sample space.

The experiment is to randomly select a human and measure his or her length. Identify the type of the sample space.

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. Finite sample space

  2. Continuous sample space

  3. Infinite discrete sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

The experiment is to randomly select a human and measure his or her length. 

Depending of how far reaching our means of selection is it is possible to consider a sample space of about $6.6$ billion humans inhabiting the planet Earth. 
In this case, the height of the selected person becomes a random variable. 
However, it is also possible to consider the sample space consisting of all possible values of height measurements of the world population. 
The tallest man ever measured lived in the United States and had a height of $272$ cm $ (8'11'')$. The height of the shortest person is more difficult to determine. Zero is clearly the low bound, but, for a living adult, it may be safely raised to, say, $40 $ cm. 
This suggests a sample space which is a line segment $[40,272]$ in centimeters. 
While at all times the human population is discrete, we may assume that in some height range near the normal average, all possible heights are realized making a continuous classification. 
We know that, "A continuous sample space is one which takes values in one or more intervals."
Thus the sample space is a continuous sample space.

Given a circle of radius $R$, the experiment is to randomly select a chord in that circle. Identify the type of the sample space.

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. Finite sample space

  2. Continuous sample space

  3. Infinite discrete sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

Given a circle of radius $R$, the experiment is to randomly select a chord in that circle.

There are many ways to accomplish such a selection. 

However the sample space is always the same:

$\{AB: A \:and\: B\: are\: points\: on\: a\: given\: circle\}$.

One natural random variable defined on this space is the length of the chord.

Here the length of the chord takes more values depending on the point we choose in the circle.

We know that, "A continuous sample space is one which takes values in one or more intervals."

Therefore, this is a continuous sample space.

Sample space for experiment in which a dice is rolled is

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $4$

  2. $8$

  3. $12$

  4. None of these

Show answer
Correct Option: D
Explanation:

A dice is rolled.

Thus $S={1,2,3,4,5,6}$
$\Rightarrow n(S)=6$.
That is, sample space for experiment in which a dice is rolled is $6$.
Since $6$ is not listed in the given options we choose D.

Choosing a birthdate is an example of .........

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. Infinite discrete sample space

  2. Finite sample space

  3. Continuous sample space

  4. None of these

Show answer
Correct Option: B
Explanation:

Let us see the sample space for choosing a birthdate.

A person can be born in any date of a month and a month has maximum of $30$ or $31$ days.
Therefore, $S={1,2,3,4,5,6,7,8,9,10,11,12,13,...,30,31}$ which is a finite set.
Thus choosing a birthdate is an example of finite sample space.

Sample space for experiment in which two coins are tossed is

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $8$

  2. $4$

  3. $2$

  4. None of these

Show answer
Correct Option: B
Explanation:

Two coins are tossed.

Number of outcomes in sample space when two coins are tossed is given by $2^2=4$.
Hence, sample space for experiment in which two coins are tossed is $4$.

In a construction job, following are some probabilities given:
Probability that there will be strike is $0.65$, probability that the job will be completed on time if there is no strike is $0.80$, probability that the job will be completed on time if there is strike is $0.32$. Determine probability that the construction job will get complete on time.

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $0.438$

  2. $0.538$

  3. $0.488$

  4. None of these

Show answer
Correct Option: C
Explanation:

Let $A$ be the event that the construction job is completed on time and $B$ be the event that there is strike.


$\Rightarrow P(B)=0.65$

Hence probability that there will be no strike $=P(B')$
                                                                            $=1-P(B)$
                                                                            $=1-0.65$
                                                                            $=0.35$

$\therefore P(B')=0.35$

By the Law of Total Probability we have $P(A)=P(B) \times P(A|B)+P(B') \times P(A|B')$

Given, $P(construction : job: is: completed: with: no: strike)=P(A|B)=0.80$ 
and $P(construction : job: is: completed: with:  strike)=P(A|B')=0.32$

$\therefore P(A)=0.65 \times 0.80+0.35 \times 0.32=0.488$

Hence the probability that the construction job will get complete on time is $0.488$

There are three boxes, each containing a different number of light bulbs. The first box has 10 bulbs, of which four are dead, the second has six bulbs, of which one is dead, and the third box has eight bulbs of which three are dead. What is the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes?

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $\dfrac{115}{330}$

  2. $\dfrac{113}{360}$

  3. $\dfrac{113}{330}$

  4. None of these

Show answer
Correct Option: B
Explanation:

Let $𝐴 _1, 𝐴 _2, 𝐴 _3$ denotes the events of selecting bulbs from bags $1,2: and: 3$ respectively. 


Let $𝐵$ denotes the event the bulbs selected are dead.

$ 𝑃(𝐴 _1) = 𝑃(𝐴 _2)  = 𝑃(𝐴 _3)  = \dfrac{1}{3} $

Also $P(B|A _1)=\dfrac{4}{10}, P(B|A _2)=\dfrac{1}{6}, P(B|A _3)=\dfrac{3}{8}$

By law of total probability,

$P(B)=P(A _1)P(B|A _1)+P(A _2)P(B|A _2)+P(A _3)P(B|A _3)$

Substituting the values we get,

$P(B)=\dfrac{1}{3} \times \dfrac{4}{10}+ \dfrac{1}{3} \times \dfrac{1}{6}+ \dfrac{1}{3} \times \dfrac{3}{8}$

$\Rightarrow P(B)=\dfrac{113}{360}$

Thus the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes is $\dfrac{113}{360}$.


Suppose that two factories supply light bulbs to the market. Factory X's bulbs work for over $5000$ hours in $99\%$ of cases, whereas factory Y's bulbs work for over $5000$ hours in $95\%$ of cases. It is known that factory X supplies $60\%$ of the total bulbs available. What is the chance that a purchased bulb will work for longer than $5000$ hours?

maths probability - iii theorem of total probability bayes theorem probability and probability distribution

  1. $\dfrac{876}{1000}$

  2. $\dfrac{544}{1000}$

  3. $\dfrac{974}{1000}$

  4. None of these

Show answer
Correct Option: C
Explanation:
Let $X$ be the event "comes from factory $X$" and $Y$ be the event "comes fom factory $Y$ " and Let $H$ be the event "works over $5000$ hours." 

Therefore $P(X)=60 \%=0.60 \Rightarrow P(Y)=1-0.60=0.40$

Given that, Factory $X's$ bulbs work for over $5000$ hours in $99 \%$ of cases.

$\therefore P(H|X)=99 \%=0.99$

Also given, factory Y's bulbs work for over $$000$5000$ hours in %$95\%$  of cases.

$\therefore P(H|Y)=95 \%=0.95$

Then by the Law of Total Probability we have

 $P(H) = P (H | X) P(X) + P (H | Y ) P(Y ) $

            $= (.99) (.6) + (.95) (.4)$

            $ = .974$


Thus $P(H)=0.974=\dfrac{974}{1000}$.


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