Probability Test

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Probability Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If $$P(A\cap B)=7/10$$ and $$P(B)=17/20$$, where $$P$$ stands for probability then $$P(A|B)$$ is equal tp

A
$$\cfrac{7}{8}$$

B
$$\cfrac{17}{20}$$

C
$$\cfrac{14}{17}$$

D
$$\cfrac{1}{8}$$

Solution

Using conditional property formula $$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$$

$$=\dfrac{\dfrac {7}{10}}{\dfrac {17}{20}}$$, substitute the given values

$$=\dfrac{14}{17}$$,
Question 2
1 / -0

A box contains $$6$$ red marbles numbers from $$1$$ through $$6$$ and $$4$$ white marbles $$12$$ through $$15$$. Find the probability that a marble drawn 'at random' is white and odd numbered

A
$$5$$

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

C
$$6$$

D
$$\dfrac {1}{6}$$

Solution

Number of white marbles are $$4$$ namely, $$12,13,14,15$$
Now, the odd numberd marbles are $$13, 15$$, two odd numbered white marbles
Total number of marbles are $$6$$ red marbles $$+$$ $$4$$ white marbles $$=10$$ marbles
Hence, $$P=\dfrac { Number\quad of\quad white\quad and\quad odd\quad numbered\quad marbles }{ Total\quad marbles }= \dfrac { 2 }{ 10 } \quad =\dfrac { 1 }{ 5 } $$
Question 3
1 / -0

Let $$A$$ and $$B$$ be two events with $$P(A^{C}) = 0.3, P(B) = 0.4$$ and $$P(A\cap B^{C}) = 0.5$$. Then, $$P(B|A\cup B^{C})$$ is equal to

A
$$\dfrac {1}{4}$$

B
$$\dfrac {1}{3}$$

C
$$\dfrac {1}{2}$$

D
$$\dfrac {2}{3}$$

Solution

Given, $$P(A^{C}) = 0.3, P(B) = 0.4$$ and $$P(A\cap B^{C}) = 0.5$$

$$\implies P(A)=1-0.3=0.7, P(B)=1-0.4=0.6$$

Now,
$$P\left (\dfrac {B}{A\cup B^{C}}\right ) = \dfrac {P(B\cap (A\cup B^{C}))}{P(A\cup B^{C})}$$

$$= \dfrac {P(A\cap B)}{P(A) + P(B^{C}) - P(A\cap B^{C})}$$

$$= \dfrac {P(A) - P(A\cap B^{C})}{P(A) + P(B^{C}) - P(A\cap B^{C})}$$

$$= \dfrac {0.7 - 0.5}{0.7 + 0.6 - 0.5} = \dfrac {0.2}{0.8} = \dfrac {1}{4}$$
Question 4
1 / -0

If $$E_1$$ denotes the events of coming sum $$6$$ in throwing two dice and $$E_2$$ be the event of coming $$2$$ in any one of the two, then $$P\left (\dfrac {E_2}{E_1}\right)$$ is

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

B
$$\dfrac {4}{5}$$

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

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

Solution

$$E_1$$ can occur as $$\{(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)\}$$
$$E_2$$ as $$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (6, 2), (5, 2), (4, 2), (3, 2), (1, 2)\}$$
$$\therefore P\left (\dfrac {E_2}{E_1}\right)=$$ Probability of $$E_2$$ when $$E_1$$ has occurred
$$=\dfrac {2}{5}$$
Question 5
1 / -0

The adjoining figure is a map of part of a city: the small rectangles are blocks and the spaces in between streets. Each morning a student walks from intersection A to intersection B, always walking along streets shown, always going east or south. For variety, at each intersection where he has a choice, he choose with probability 1/2 (independent of all other choices) whether to go east or south. Find the probability that, on any given morning, he walks through intersection C.

A
$$\dfrac{11}{23}$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{4}{7}$$

D
None of these

Solution

Probability of the student walking through East or south $$=\dfrac{1}{2}.$$
If the boy passes through intersection $$C,$$ then he would choose a path east of the town and walk towards west So, as he walk South he would be passing through intersection $$C$$ to in reach intersection $$D.$$
$$\therefore$$ Probability of the choosing the odd on $$7.$$
$$\Rightarrow$$ Days of a work $$=\dfrac{1}{7}\times\dfrac{1}{2}=\dfrac{1}{14}.$$
Hence, the answer is none of these.
Question 6
1 / -0

The following is probabilty distribution of r.v X.
x 1 2 3 4 5 6
p(x) $$\frac{k}{6}$$ $$\frac{k}{6}$$ $$\frac{k}{6}$$ $$\frac{k}{6}$$ $$\frac{k}{6}$$ $$\frac{k}{6}$$
then value of k is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$
Solution
We know that, Suppose a random variable $$X$$ may take $$k$$ different values, with the probability that $$X=x_i$$ defined to be $$P(X=x_i)=p_i$$. The probabilities $$p_i$$ must satify the following:
1. $$0\leq p_i\leq1$$ for each $$i$$.
2. $$p_1+p_2+...+p_k=1$$ that is $$\sum p_i=1$$
Thus we have $$\sum p_i=1$$

$$\Rightarrow p_1+p_2+p_3+p_4+p_5+p_6=1$$

$$\Rightarrow \dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}+\dfrac{k}{6}=1$$

$$\Rightarrow \dfrac{6k}{6}=1$$

$$\Rightarrow k=1$$
Question 7
1 / -0

The variable which takes some specific values is called

A
discrete random variable

B
continuous random variable

C
both A and B

D
none

Solution

A random variable is a variable whose value is a numerical outcome of a random experiment.

A discrete random variable $$X$$ has a countable number of possible values.

That is, a random variable that takes some specific values is called discrete random variable.
Question 8
1 / -0

The probability distribution of random variable X: number of heads , when a fair coin is tossed twice is given by
$$x$$ $$0$$ $$1$$ $$2$$
p(x) $$p_1$$ $$p_2$$ $$p_3$$
then

A
$$\sum p_i=0$$

B
$$\Pi p_i=1$$

C
$$\sum p_i=1$$

D
$$\sum p_i=3$$

Solution

We know that, the probability distribution of a random variable is a list of probabilities associated with each of its possible values.

Suppose a random variable $$X$$ may take $$k$$ different values, with the probability that $$X=x_i$$ defined to be$$P(X=x_i)=p_i$$.

The probabilities $$p_i$$ must satisfy the following:

$$0\leq p_i \leq 1$$ for each $$i$$
$$p_1+p_2+...+p_k=1. \:That \:is\: \sum p_i=1$$

Thus if the probability distribution of random variable $$X$$: number of heads , when a fair coin is tossed twice is given then $$\sum p_i=1$$
Question 9
1 / -0

To define probability disribution function we assign to each variable

A
the respective probabilities

B
the specific random values

C
some integers

D
none

Solution

We know that, the probability distribution function of a random variable is a list of probabilities associated with each of its possible values.

Thus to define probability distribution function we assign to each variable the respective probabilities.
Question 10
1 / -0

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

A
$$\dfrac{2}{3}$$

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

C
$$\dfrac{1}{2}$$

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

Solution

Consider $$2$$ coins,
In the care of unbiased coin, probability of getting head $$=P(E_1)=\left( \dfrac { { 1 } }{ { 2 } } \right) $$
In the care of biased coin, probability of getting head $$=P(E_2)=\left( \dfrac { { 3 } }{ { 4 } } \right) $$
The probability of getting a biased or unbiased coin $$=P\left( \dfrac { A }{ E_1} \right)=P\left( \dfrac { A }{ E_2 } \right)=\dfrac{1}{2}$$
$$\Rightarrow A=$$ Probability that unbiased coin is selected.
$$\Rightarrow P\left( \dfrac { E }{ E_1 } \right)=\dfrac { P\left( { E }_{ 1 } \right) P\left( \dfrac{A}{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) P\left( \dfrac{A}{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) P\left( \dfrac{A}{ E }_{ 2 } \right) } $$
                        $$=\dfrac { \dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } }{ \dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \times \dfrac { 3 }{ 4 } } $$
                        $$=\dfrac { \dfrac { 1 }{ 4 } }{ \dfrac { 1 }{ 4 } +\dfrac { 3 }{ 8 } } $$
                        $$=\dfrac { \dfrac { 1 }{ 4 } }{ \dfrac { 5 }{ 8 } } $$
                        $$=\dfrac { 2 }{ 5 }.$$
Hence, the answer is $$\dfrac { 2 }{ 5 }.$$

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x	1	2	3	4	5	6
p(x)	$$\frac{k}{6}$$	$$\frac{k}{6}$$	$$\frac{k}{6}$$	$$\frac{k}{6}$$	$$\frac{k}{6}$$	$$\frac{k}{6}$$

$$x$$	$$0$$	$$1$$	$$2$$
p(x)	$$p_1$$	$$p_2$$	$$p_3$$

Probability Test - 27

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Probability Test - 27

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SHARING IS CARING

Question 1

$$\cfrac{7}{8}$$

$$\cfrac{17}{20}$$

$$\cfrac{14}{17}$$

$$\cfrac{1}{8}$$

Solution

Question 2

$$5$$

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

$$6$$

$$\dfrac {1}{6}$$

Solution

Question 3

$$\dfrac {1}{4}$$

$$\dfrac {1}{3}$$

$$\dfrac {1}{2}$$

$$\dfrac {2}{3}$$

Solution

Question 4

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

$$\dfrac {4}{5}$$

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

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

Solution

Question 5

$$\dfrac{11}{23}$$

$$\dfrac{1}{2}$$

$$\dfrac{4}{7}$$

None of these

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 7

discrete random variable

continuous random variable

both A and B

none

Solution

Question 8

$$\sum p_i=0$$

$$\Pi p_i=1$$

$$\sum p_i=1$$

$$\sum p_i=3$$

Solution

Question 9

the respective probabilities

the specific random values

some integers

none

Solution

Question 10

$$\dfrac{2}{3}$$

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

$$\dfrac{1}{2}$$

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

Solution

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