Probability Test 12

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Probability Test 12

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Question 1
1 / -0

If we throw a dice, then the sample space, $$S = {1, 2, 3, 4, 5, 6}$$. Now the event of $$3$$ appearing on the dice is simple and given by

A
$$E = {2, 3}$$

B
$$E = {1, 2, 3}$$

C
$$E = {3}$$

D
$$E = {1, 3}$$

Solution

If we throw a dice, then the sample space, $$S = {1, 2, 3, 4, 5, 6}$$. Now the event of 3 appearing on the dice is simple and given by $$E = {3}$$
If there be only one element of the sample space in the set representing an event, then this event is called a simple or elementary event.
Question 2
1 / -0

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are $$\displaystyle \frac { 1 }{ 4 } ,\frac { 1 }{ 2 } $$ and $$\displaystyle \frac { 5 }{ 8 } $$, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A
$$\displaystyle \frac { 11 }{ 8 } $$

B
$$\displaystyle \frac { 7 }{ 8 } $$

C
$$\displaystyle \frac { 9 }{ 64 } $$

D
$$\displaystyle \frac { 5 }{ 64 } $$

E
$$\displaystyle \frac { 3 }{ 64 } $$

Solution

P(Xavier will solve)$$=\dfrac{1}{4}$$
P(Yvonne will solve)$$=\dfrac{1}{2}$$
P(Zelda will NOT solve) $$= 1- \dfrac{5}{8} = \dfrac{3}{8}.$$

Now, we need to multiply all this Ps to find an answer:
$$p= (\dfrac{1}{4})*(\dfrac{1}{2})*(\dfrac{3}{8}) = \dfrac{3}{64}.$$
Question 3
1 / -0

A coin is tossed $$100$$ times with the following frequencies: Head : $$20$$. Find the probability for event having heads only.

A
$$0.2$$

B
$$0.5$$

C
$$0.65$$

D
$$1.5$$

Solution

Since the coin is tossed $$100$$ times, the total number of trials is $$100$$.
Let us call the events of getting a head and of getting a tail as $$X$$ and $$Y$$, respectively. Then, the number of times $$X$$ happens, i.e., the number of times a head come up, is $$20$$.
$$P(X) =$$ $$\dfrac{\text{Number space of space heads}}{\text{Total space number space of space trials}}$$
$$= \dfrac{20}{100}= \dfrac{1}{5} = 0.2$$
Question 4
1 / -0

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

A
numerical complementary events

B
equal compulsory events

C
complementary events

D
compulsory events

Solution

Difference between sample space and subset of sample space is considered complementary events.
Complementary events are two outcomes of an event that are the only two possible outcomes. This is like flipping a coin and getting heads or tails. ... Rolling a die and getting a 1 or 2 are not complementary since there are other outcomes that may happen (3, 4, 5, or 6).
Question 5
1 / -0

Two coins are available, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability $$\dfrac {3}{4}$$. Given that the outcome is head the probability that the two-headed coin was chosen, is

A
$$\dfrac {3}{5}$$

B
$$\dfrac {2}{5}$$

C
$$\dfrac {1}{5}$$

D
$$\dfrac {2}{7}$$

Solution

Let $$F$$ denotes fair coin.
$$T$$ denotes two headed,

$$H$$ denotes head occurs

$$\therefore P(F) = \dfrac {3}{4}, P(T) = 1 - \dfrac {3}{4} = \dfrac {1}{4}$$

$$\therefore P\left (\dfrac {T}{H}\right ) = \dfrac {P\left (\dfrac {H}{T}\right )\cdot P(T)}{P\left (\dfrac {H}{T}\right )\cdot P(T) + P\left (\dfrac {H}{F}\right )\cdot P(F)}$$ (By Baye's theorem)

$$= \dfrac {1\cdot \dfrac {1}{4}}{1\cdot \dfrac {1}{4} + \dfrac {1}{2}\cdot \dfrac {3}{4}} = \dfrac {2}{5}$$
Question 6
1 / -0

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

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

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

C
Both are correct

D
None of these

Solution

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.
Question 7
1 / -0

A bag contains three red balls and two blue balls. We remove successively, and with replacement, two balls and observe their colour. What is the probability of "to extract a red ball and a blue one, without taking into account the order"?

A
$$\dfrac{12}{25}$$

B
$$\dfrac{13}{25}$$

C
$$\dfrac{12}{23}$$

D
None of these

Solution

Given:
Total number of balls in a bag = 3 red balls + 2 blue balls = 5
Let's consider the events: R= "to remove a red ball", A=R'= "to remove a blue ball.
Our sample space is: (RR, RA, AR, AA}, and the events that we are interested in are RA and AR.
We represent our problem in a tree as shown in the figure
Using the law of total probability,
$$P(S)=P(R)\times P(A/R)+P(A)\times P(R/A)=\dfrac 35\times \dfrac 25+\dfrac 25\times \dfrac 35=\dfrac {12}{25}$$
is the required probability.
Question 8
1 / -0

Let $$S$$ be a set containing $$n$$ elements and we select two subsets $$A$$ and $$B$$ of $$S$$ at random, then the probability that $$A\cup B=S$$ and $$A\cap B=\phi$$, is.

A
$$2^n$$

B
$$n^2$$

C
$$\dfrac 1n$$

D
$$\dfrac {1}{2^n}$$

Solution

Given that $$S$$ contains $$n$$ elements and two sets $$A$$ and $$B$$ are selected.
Two sets $$A$$ and $$B$$ can be selected in $$2^{n}$$ ways.
The number of ways of selecting two sets such that their union is $$S$$ and intersection is nullset is $$1$$.
Therefore the probability is $$\frac{1}{2^{n}}$$.
So the correct option is $$D$$.
Question 9
1 / -0

There are two coins, one unbiased with probability $$\cfrac{1}{2}$$ of getting heads and the other one is biased with probability $$\cfrac{3}{4}$$ of getting heads. A coin is selected at random and tossed. It shows heads up. Then, the probability that the unbiased coin was selected is

A
$$\cfrac{2}{3}$$

B
$$\cfrac{3}{5}$$

C
$$\cfrac{1}{2}$$

D
$$\cfrac{2}{5}$$

Solution

Let $$E\rightarrow$$ Event of head showing up
$${E}_{1}\rightarrow$$ Event of biased coin chosen
$${E}_{2}\rightarrow$$ Event of unbiased coin chosen
Now, $$P({E}_{2})=\cfrac{1}{2}$$ and $$P({E}_{1})=\cfrac{1}{2}$$
Also, $$P(\cfrac{E}{{E}_{2}})=\cfrac{1}{2}$$ and $$P(\cfrac{E}{{E}_{1}})=\cfrac{3}{4}$$
(by conditional probability)
By Baye's theorem
$$P\left( \cfrac { { E }_{ 2 } }{ E } \right) =\cfrac { P({ E }_{ 2 }).P\left( \cfrac { E }{ { E }_{ 2 } } \right) }{ P({ E }_{ 2 }).P\left( \cfrac { E }{ { E }_{ 2 } } \right) +P({ E }_{ 1 }).P\left( \cfrac { E }{ { E }_{ 1 } } \right) } $$
$$=\cfrac { \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } }{ \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } \times \cfrac { 3 }{ 4 } } =\cfrac { 2 }{ 5 } $$
Question 10
1 / -0

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?

A
$$0.60$$

B
$$0.40$$

C
$$0.50$$

D
None of these

Solution

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.

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Probability Test 12

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Probability Test 12

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Question 1

$$E = {2, 3}$$

$$E = {1, 2, 3}$$

$$E = {3}$$

$$E = {1, 3}$$

Solution

Question 2

$$\displaystyle \frac { 11 }{ 8 } $$

$$\displaystyle \frac { 7 }{ 8 } $$

$$\displaystyle \frac { 9 }{ 64 } $$

$$\displaystyle \frac { 5 }{ 64 } $$

$$\displaystyle \frac { 3 }{ 64 } $$

Solution

Question 3

$$0.2$$

$$0.5$$

$$0.65$$

$$1.5$$

Solution

Question 4

numerical complementary events

equal compulsory events

complementary events

compulsory events

Solution

Question 5

$$\dfrac {3}{5}$$

$$\dfrac {2}{5}$$

$$\dfrac {1}{5}$$

$$\dfrac {2}{7}$$

Solution

Question 6

It 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

Both are correct

None of these

Solution

Question 7

$$\dfrac{12}{25}$$

$$\dfrac{13}{25}$$

$$\dfrac{12}{23}$$

None of these

Solution

Question 8

$$2^n$$

$$n^2$$

$$\dfrac 1n$$

$$\dfrac {1}{2^n}$$

Solution

Question 9

$$\cfrac{2}{3}$$

$$\cfrac{3}{5}$$

$$\cfrac{1}{2}$$

$$\cfrac{2}{5}$$

Solution

Question 10

$$0.60$$

$$0.40$$

$$0.50$$

None of these

Solution

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