Tag: probability and probability distribution

Questions Related to probability and probability distribution

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

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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.

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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

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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

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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

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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.

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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

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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

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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

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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.