Probability Test

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Probability Test - 25

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Question 1
1 / -0

If $$P(A\cup B)=\dfrac {2}{3}, P(A\cap B)=\dfrac {1}{6}$$ and $$P(A)=\dfrac {1}{3}$$, then

A
A and B are independent events

B
A and b are disjoint events

C
A and b are dependent events

D
None of these

Solution

If $$P(A\cup B)=\frac {2}{3}$$
$$P(A\cap B) = \frac {1}{6}$$
$$P(A)=\frac {1}{3}$$
$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$
$$\frac {2}{3} = \frac {1}{3} + P(B) - \frac {1}{6}$$
$$P(B) = \frac{1}{2}$$

Since, $$ P(A\cap B) = P(A).P(B)$$
Therefore, the events are independent.
Question 2
1 / -0

A bag contains $$2$$ red, $$3$$ green and $$2$$ blue balls. $$1$$ ball is to be drawn randomly. What is the probability that the ball drawn is not blue?

A
$$\dfrac {5}{7}$$

B
$$\dfrac {10}{21}$$

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

D
$$\dfrac {11}{21}$$

Solution

There are $$2$$ red, $$3$$ green, and $$2$$ blue balls
Total number of possibilities$$= 7$$
The possibility of the ball not being blue$$=5$$
$$\therefore$$ Probability = $$\dfrac{5}{7}$$
Question 3
1 / -0

A bag contains $$5$$ red balls and $$8 $$ blue balls. It also contains $$4$$ green and $$7$$ black balls. If a ball is drawn at random, then find the probability that it is not green.

A
$$\dfrac{5}{6}$$

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

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

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

Solution

Toatal number of favourable cases that a ball is not green $$= 20$$
Total number of cases$$ = 24$$
Probability = $$\dfrac{20}{24}$$ = $$\dfrac{5}{6}$$
Question 4
1 / -0

The probability that a card drawn from a pack of $$52$$ cards will be a diamond or a king is -

A
$$\displaystyle \frac{4}{52}$$

B
$$\displaystyle \frac{4}{13}$$

C
$$\displaystyle \frac{17}{52}$$

D
$$\displaystyle \frac{2}{13}$$

Solution

$${\textbf{Step-1: Find sample space and favorable event.}}$$

$${\text{Let S be the sample space.}}$$

$$\Rightarrow n(S)=52$$

$${\text{Let A be the event getting a diamond or a king.}}$$

$${\text{There are 13 diamond cards including its king and other 3 king cards.}}$$

$${\text{so total favorable cards are,}}$$

$$\Rightarrow n(A)=13 + 3= 16$$

$${\textbf{Step-2: Find probability.}}$$

$$\Rightarrow P(A)= \dfrac{n(A)}{n(S)}$$

$$\Rightarrow P(A)= \dfrac{16}{52}$$

$$\Rightarrow P(A)= \dfrac{4}{13}$$

$${\textbf{Hence, option B is correct.}}$$
Question 5
1 / -0

If A and B are two events such that $$P(A\cup B)=P(A\cap B)$$, then the incorrect statement amongst the following statements is :

A
A and B are equally likely

B
$$P(A\cap B')=0$$

C
$$P(A'\cap B)=0$$

D
$$P(A)+P(B)=1$$

Solution

$$P(A\cup B)=P(A\cap B)$$
$$\Rightarrow P(A)+P(B)-P(A\cap B)= P(A\cap B)$$
$$\Rightarrow P(A)+P(B) = 2P(A\cap B)$$
Question 6
1 / -0

If events $$A$$ and $$B$$ are independent and $$P(A)=0.15, P(A\cup B)=0.45,$$ then $$P(B)$$ is:

A
$$\dfrac {6}{13}$$

B
$$\dfrac {6}{17}$$

C
$$\dfrac {6}{19}$$

D
$$\dfrac {6}{23}$$

Solution

$$P(A\cap B)=P(A)\times P(B)$$
$$=0.15\times P(B)$$
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$0.45=0.15+P(B)-(0.15)\times P(B)$$
$$=0.15+P(B)(1-0.15)$$
$$=0.15+0.85P(B)$$
$$\therefore 0.85P(B)=0.45-0.15=0.30$$
$$\therefore P(B)=\dfrac {0.30}{0.85}=\dfrac {30}{85}=\dfrac {6}{17}$$
Question 7
1 / -0

A set $$A$$ is containing $$n$$ elements. A subset $$P$$ of $$A$$ is chosen at random. The set is reconstructed by replacing the elements of $$P$$. A subset $$Q$$ of $$A$$ is again chosen at random. The probability that $$P$$ and $$Q$$ have no common elements is:

A
$$\left(\dfrac{5}{6}\right)^n$$

B
$$\left (\dfrac {3}{4}\right )^n$$

C
$$\left(\dfrac {3}{5}\right)^n$$

D
$$\left(\dfrac{2}{3}\right)^n$$

Solution

Set $$A$$ has $$n$$ elements. Number of subsets of $$A$$ = $$2^n$$.
Number of ways of choosing a subset from $$A$$ = $${^{2^n}}C_{1}$$ ways = $$2^n$$ ways.
So, number of ways of selecting two subsets from $$A$$ = $$2^n$$ $$ \times$$ $$2^n$$ = $$ 4^n$$ ways. -----------------(1)

$$P$$ is a subset of $$A$$.
Let set $$P$$ has $$r$$ elements.
These $$r$$ elements are chosen from $$n$$ elements of set $$A$$ ($$r$$ varies from $$0$$ to $$n$$).
This can be done in $$\bf {^n}C_{r}$$ ways.

Now given that $$P$$ and $$Q$$ has no common elements, $$P \cap Q = \emptyset$$.
Hence the elements of $$Q$$ has to be chosen from $$n-r$$ elements.
This can be done in $$^{n-r}C_{0} + ^{n-r}C_{1} + ................. + ^{n-r}C_{n-r}$$ ways = $$ 2^{n-r}$$ ways.

Hence $$P$$ and $$Q$$ can be chosen in $$ ^{n}C_{r}\times 2^{n-r}$$ ways.
Now $$r$$ can vary from $$0$$ to $$n$$.

Hence total number of ways in which $$P$$ and $$Q$$ can be chosen such that they dont have any common elements
$$= \displaystyle \sum_{r=0}^{n} $$ $$^{n}C_{r} \times $$$$2^{n-r}$$

$$= \displaystyle \sum_{r=0}^{n} $$$$^{n}C_{r} \times $$$$1^{r}\times 2^{n-r}$$

$$=(1+2)^{n}= 3^n$$ -------------------------(2) (Using binomial expansion)

Therefore, probability that $$P$$ and $$Q$$ have no common elements
$$=\dfrac{3^n}{4^n} = {\big (\dfrac{3}{4}\big )}^n$$ (From (1) and (2))
Question 8
1 / -0

If events $$A$$ and $$B$$ are independent and $$P(A)=0.4, P(A\cup B)=0.6$$, then $$P(B)=$$

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

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

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

D
None of these

Solution

Given $$A$$ and $$B$$ are independent events, $$P(A)=0.4, P(A \cup B)=0.6$$. We have to find $$P(B)$$.
Since $$A$$ and $$B$$ are independent events, $$P(A \cap B)=P(A) \cdot P(B)$$.
We know that, $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
$$\Rightarrow P(A \cup B)=P(A)+P(B)-P(A) \cdot P(B)$$
Substitute the values we get
$$0.6=0.4+P(B)-0.4 \times P(B) $$
$$\Rightarrow 0.6-0.4=(1-0.4)P(B)$$
$$\Rightarrow 0.6P(B)=0.2$$
$$\Rightarrow P(B)=\dfrac{0.2}{0.6}=\dfrac{2}{6}=\dfrac{1}{3}$$
That is $$P(B)=\dfrac{1}{3}$$
Question 9
1 / -0

An unbiased normal coin is tossed n times. Let $$E_1$$: event that both heads and tails are present in n tosses. $$E_2$$:event that the coin shows up heads at most once. The value of n for which $$E_1$$ and $$E_2$$ are independent is

A
1

B
2

C
3

D
4

Solution

Let's denote head by 'H' and tails by 'T'
$$P\left( { E }_{ 1 } \right) =1-\left\{ P\left( all\quad Ts \right) +P\left( all\quad Hs \right) \right\} \\ \quad \quad =1-\left\{ \dfrac { 1 }{ { 2 }^{ n } } +\dfrac { 1 }{ { 2 }^{ n } } \right\} =1-\dfrac { 1 }{ { 2 }^{ n-1 } } $$
so $${ E }_{ 2 }=\left( n-1 \right) T\quad and\quad 1H\quad or\quad n\quad T\quad \\ \Rightarrow P\left( { E }_{ 2 } \right) =\left( \begin{matrix} n \\ 1 \end{matrix} \right) { 0.5 }^{ 1 }\times { 0.5 }^{ n-1 }\quad +\quad { 0.5 }^{ n }\\ \Rightarrow P\left( { E }_{ 2 } \right) =\dfrac { n+1 }{ { 2 }^{ n } } $$
It can be seen very clearly that $${ E }_{ 1 }\cap { E }_{ 2 }=\left( n-1 \right) T\quad and\quad 1H$$
$$\Rightarrow P\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =\left( \begin{matrix} n \\ 1 \end{matrix} \right) { 0.5 }^{ 1 }\times { 0.5 }^{ n-1 }\\ =\dfrac { n }{ { 2 }^{ n } } $$
now for independence $$P\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =P\left( { E }_{ 1 } \right) P\left( { E }_{ 2 } \right) \\ \Rightarrow \dfrac { n }{ { 2 }^{ n } } =\left( 1-\dfrac { 1 }{ { 2 }^{ n-1 } } \right) \left( \dfrac { n+1 }{ { 2 }^{ n } } \right) \\ \Rightarrow n\times { 2 }^{ n-1 }=\left( n+1 \right) \left( { 2 }^{ n-1 }-1 \right) \\ \Rightarrow { 2 }^{ n-1 }-n-1=0\\ \Rightarrow n=3$$
Question 10
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Events A and C are independent. If the probability relating A, B and C are $$P(A)=1/5, P(B)=1/6;$$ $$P (A \cap C) = 1/20. P(B \cup C) = 3/8$$. Then

A
events B and C are independent

B
events B and C are mutually exclusive

C
events B and C are neither independent nor mutually exclusive

D
events B and C are equiprobable

Solution

Given $$\displaystyle P\left( A \right) =\frac { 1 }{ 5 } ,P\left( B \right) =\frac { 1 }{ 6 } ,P\left( A\cap C \right) =\frac { 1 }{ 20 } \quad $$and $$\displaystyle \quad P\left( B\cup C \right) =\frac { 3 }{ 8 } $$
since A and C are independent
$$\Rightarrow \displaystyle P\left( A\cap C \right) =P\left( A \right) P\left( C \right) =\frac { 1 }{ 20 } \\ \displaystyle \Rightarrow P\left( C \right) =\frac { 1 }{ 20 } \times 5=\frac { 1 }{ 4 } \\ \Rightarrow P\left( C \right) =\dfrac { 1 }{ 4 } \\ Also \displaystyle \quad P\left( B\cup C \right) =\frac { 3 }{ 8 } \quad \quad \Rightarrow P\left( B \right) +P\left( C \right) -P\left( B\cap C \right) =\frac { 3 }{ 8 } \\ \displaystyle \Rightarrow P\left( B\cap C \right) =\frac { 1 }{ 6 } +\frac { 1 }{ 4 } -\frac{3}{8} =\frac { 1 }{ 24 } \\ \displaystyle \Rightarrow P\left( B\cap C \right) =P\left( B \right) P\left( C \right) =\frac { 1 }{ 24 } $$
Which means events $$B$$ and $$C$$ are independent

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Probability Test - 25

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Probability Test - 25

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Question 1

A and B are independent events

A and b are disjoint events

A and b are dependent events

None of these

Solution

Question 2

$$\dfrac {5}{7}$$

$$\dfrac {10}{21}$$

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

$$\dfrac {11}{21}$$

Solution

Question 3

$$\dfrac{5}{6}$$

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

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

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

Solution

Question 4

$$\displaystyle \frac{4}{52}$$

$$\displaystyle \frac{4}{13}$$

$$\displaystyle \frac{17}{52}$$

$$\displaystyle \frac{2}{13}$$

Solution

Question 5

A and B are equally likely

$$P(A\cap B')=0$$

$$P(A'\cap B)=0$$

$$P(A)+P(B)=1$$

Solution

Question 6

$$\dfrac {6}{13}$$

$$\dfrac {6}{17}$$

$$\dfrac {6}{19}$$

$$\dfrac {6}{23}$$

Solution

Question 7

$$\left(\dfrac{5}{6}\right)^n$$

$$\left (\dfrac {3}{4}\right )^n$$

$$\left(\dfrac {3}{5}\right)^n$$

$$\left(\dfrac{2}{3}\right)^n$$

Solution

Question 8

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

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

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

None of these

Solution

Question 9

1

2

3

4

Solution

Question 10

events B and C are independent

events B and C are mutually exclusive

events B and C are neither independent nor mutually exclusive

events B and C are equiprobable

Solution

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