Probability Test

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Probability Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

Let A and E be any two events with positive probabilities :
Statement - 1 : $$P \left (\displaystyle \frac{E}{A} \right) \geq P \left (\displaystyle \frac{A}{E} \right ) P(E)$$
Statement - 2 : $$P \left (\displaystyle \frac{A}{E}\right ) \geq P(A\cap E)$$

A
Both the statement are true

B
Both the statement are false

C
Statement - 1 is true, Statement - 2 is false

D
Statement - 1 is false, Statement - 2 is true.

Solution

$$\displaystyle P(E|A)=\frac { P(E\cap A) }{ P(A) } $$

$$\displaystyle P(A|E)=\frac { P(E\cap A) }{ P(E) } $$

$$0\le P(E)\le 1$$

$$\displaystyle P(A|E)\ge P(E\cap A)$$

$$\displaystyle P(E|A)=\frac { P(A|E)P(E) }{ P(A) } $$ .

$$0\le P(A)\le 1$$
So,$$\displaystyle P(E|A) \ge { P(A|E)P(E) }$$
Question 2
1 / -0

One ticket is selected at random from $$50$$ tickets numbered $$00,\ 01,\ 02, ... , 49$$. Then the probability that the sum of the digits on the selected ticket is $$8$$, given that the product of these digits is zero, equals

A
$$1/14$$

B
$$1/7$$

C
$$5/14$$

D
$$1/50$$

Solution

A $$=$$Events that sum of the digits on selected ticket is 8
$$=\{08,17,26,35,44\}$$

Therefore, total $$5$$ cases when sum of digits is $$8$$

$$\Rightarrow \mathrm{n}(\mathrm{A})=5$$

Event that product of digits is zero

$$=$$ $$\{00,01,02,03,04,05,06,07,08,09,10,20,30, 40\}$$

$$\Rightarrow \mathrm{n}(\mathrm{A \cap B})=1$$ as only $$08$$ is the number with sum $$8$$ and product zero

$$= \mathrm{n}(\mathrm{B})=14$$

$$\displaystyle \mathrm{P}(\frac{\mathrm{A}}{\mathrm{B}})=\dfrac {n(A\cap B)}{n(B)}=\frac{1}{14}$$
Question 3
1 / -0

Three numbers are chosen at random without replacement from $$\{1, 2, 3, ..., 8\}$$. The probability that their minimum is $$3$$, given that their maximum is $$6$$, is:

A
$$\displaystyle \frac{3}{8}$$

B
$$\displaystyle \frac{1}{5}$$

C
$$\displaystyle \frac{1}{4}$$

D
$$\displaystyle \frac{2}{5}$$

Solution

$$A\rightarrow$$ Maximum is 6
$$B\rightarrow$$ Minimum is 3
$$\displaystyle P(A)=\cfrac {^5C_2}{^8C_3}$$
$$\displaystyle P(B)=\cfrac {^5C_2}{^8C_3}$$
$$\displaystyle P(A\cap B)=\cfrac {^2C_1}{^8C_3}$$
$$\displaystyle P(\frac {B}{A})=\cfrac {P(A\cap B)}{P(A)}=\cfrac {^2C_1}{^5C_2}=\cfrac {2}{10}=\cfrac {1}{5}$$
Question 4
1 / -0

For three events $$A, B$$ and $$C$$, $$P$$ (Exactly one of $$A$$ or $$B$$ occurs) $$= P$$ (Exactly one of $$B$$ or $$C$$ occurs) $$=P$$(Exactly one of $$C$$ or $$A$$ occurs) $$=\displaystyle\frac{1}{4}$$and $$P$$ (All the three events occur simultaneously)$$=\displaystyle\frac{1}{16}$$. Then the probability that at least one of the events occurs, is.

A
$$\displaystyle\frac{7}{32}$$

B
$$\displaystyle\frac{7}{16}$$

C
$$\displaystyle\frac{7}{64}$$

D
$$\displaystyle\frac{3}{16}$$

Solution

$$P$$(exactly one of $$A$$ or $$B$$)$$=P(A\cup B)-P(A\cap B)=\dfrac{1}{4}=P(A)+P(B)-2P(A\cap B)$$

$$P$$(exactly one of $$C$$ or $$B$$)$$=P(C\cup B)-P(C\cap B)=\dfrac{1}{4}=P(C)+P(B)-2P(C\cap B)$$

$$P$$(exactly one of $$A$$ or $$C$$)$$=P(A\cup C)-P(A\cap C)=\dfrac{1}{4}=P(A)+P(C)-2P(A\cap C)$$
Adding all
$$2P(A)+2P(B)+2P(c)-2P(A\cap B)-2P(A\cap C)-2P(B\cap C)=\dfrac{3}{4}$$

$$P(A)+P(B)+P(c)-P(A\cap B)-P(A\cap C)-P(B\cap C)=\dfrac{3}{8}$$

$$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$$

=$$\dfrac{3}{8}+\dfrac{1}{16}=\dfrac{7}{16}$$
Question 5
1 / -0

If $$\mathrm{C}$$ and $$\mathrm{D}$$ are two events such that $$\mathrm{C}\subset \mathrm{D}$$ and $$\mathrm{P}(\mathrm{D})\neq 0$$, then the correct statement among the following is

A
$$P\left(\dfrac{C}{D}\right) =\mathrm{P}(\mathrm{C})$$

B
$$P\left(\dfrac{C}{D}\right) \geq \mathrm{P}(\mathrm{C})$$

C
$$P\left(\dfrac{C}{D}\right) <\mathrm{P}(\mathrm{C})$$

D
$$P\left(\dfrac{C}{D}\right) =\displaystyle \frac{\mathrm{P}(\mathrm{D})}{\mathrm{P}(\mathrm{C})}$$

Solution

$$\displaystyle \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{D}}\right)=\frac{\mathrm{P}(\mathrm{C}\mathrm{\cap}\mathrm{D})}{\mathrm{P}(\mathrm{D})}=\frac{\mathrm{P}(\mathrm{C})}{\mathrm{P}(\mathrm{D})}\geq\mathrm{P}(\mathrm{C})$$ (Since $$0<\mathrm{P}(\mathrm{D})\leq 1$$)
So, $$\displaystyle \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{D}}\right)\geq \mathrm{P}(\mathrm{C})$$
Hence, option 'B' is correct.
Question 6
1 / -0

Let A, B and C be three events, which are pair-wise independence and $$\overline{E}$$ denotes the complement of an event E. If $$P(A\cap B\cap C)=0$$ and $$P(C) > 0$$, then $$P[(\overline{A}\cap \overline{B})|C]$$ is equal to.

A
$$P(A)+P(\overline{B})$$

B
$$P(\overline{A})-P(\overline{B})$$

C
$$P(\overline{A})-P(B)$$

D
$$P(\overline{A})+P(\overline{B})$$

Solution

we need find $$P(\bar {A} \cap \bar {B}| {C})$$ = shaded portions in Venn Diagram

=$$P(\bar {A} \cap \bar {B}| {C}) = \dfrac{P(\bar {A} \cap \bar {B}\cap {C})}{P(C)}$$

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

=$$1-\dfrac {P(A) . P(C) + P(B).P(C)}{P(C)}$$

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

=$$P(\bar {A})- P(B)$$
Question 7
1 / -0

If $$A$$ and $$B$$ are any two events such that $$P(A) = \dfrac {2}{5}$$ and $$P(A\cap B) = \dfrac {3}{20}$$, then the conditional probability, $$P(A|(A'\cup B'))$$, where A' denotes the complement of $$A$$, is equal to:

A
$$\dfrac {5}{17}$$

B
$$\dfrac {11}{20}$$

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

D
$$\dfrac {8}{17}$$

Solution

$$P(A|(A'\cup B')) = \dfrac{P(A\cap(A'\cup B'))}{P(A' \cup B')}$$

$$P(A\cap(A'\cup B'))=P(A)-P(A\cap B)= \dfrac{2}{5}-\dfrac{3}{20}=\dfrac{5}{20}$$
$$P(A' \cup B')=1-P(A\cap B)=\dfrac{17}{20}$$

Hence, $$P(A|(A'\cup B')) =\dfrac {\dfrac{5}{20}}{\dfrac{17}{20}}=\dfrac{5}{17}$$
Question 8
1 / -0

It is given that the events A and $$B$$ are such that $$P(A)=\displaystyle \frac{1}{4},\ P(A|B)=\displaystyle \frac{1}{2}$$ and $$P(B|A)=\displaystyle \frac{2}{3}$$. Then $$P(B)$$ is

A
$$\displaystyle \frac{2}{3}$$

B
$$\displaystyle \frac{1}{2}$$

C
$$\displaystyle \frac{1}{6}$$

D
$$\displaystyle \frac{1}{3}$$

Solution

$$P(A)=\displaystyle \frac{1}{4}$$
$$P(\displaystyle A|B)=\frac{P(A\mathrm{\cap}B)}{P(B)} = \displaystyle \frac{1}{2}$$ ... (i)
$$P(\displaystyle B|A)=\frac{P(A\mathrm{\cap}B)}{P(A)}= \displaystyle

\frac{2}{3}=\frac{P(A\mathrm{\cap}B)}{1/4}= P(A\mathrm{\cap} B)=\displaystyle

\frac{1}{6}$$.
Putting the value of $$P(A\mathrm{\cap} B)$$ in (i) $$,

P(B)=2\displaystyle \times\frac{1}{6}=\frac{1}{3}$$
Question 9
1 / -0

A bag contains $$4$$ red and $$6$$ black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is

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

B
$$\dfrac {3}{4}$$

C
$$\dfrac {3}{10}$$

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

Solution

There are $$2$$ possible cases: first ball drawn is red or is black.
$$\therefore$$ P(second ball red | first ball drawn is black) $$=$$ $$\dfrac{4}{4+8}=\dfrac{1}{3}$$

$$\therefore$$ P(second ball red | first ball drawn is red) $$=$$ $$\dfrac{4+2}{6+6}=\dfrac{1}{2}$$

Hence, the probability of the second ball being red
$$=$$ P(first ball black)$$\times$$P(second ball red | first ball black)+P(first ball red)$$\times$$P(second ball red | first ball red)

$$=\dfrac{6}{10}\times\dfrac{1}{3}+\dfrac{4}{10}\times\dfrac{1}{2}$$

$$=\dfrac{2}{5}$$
This is the required probability.
Question 10
1 / -0

A box $$'A'$$ contains $$2$$ white, $$3$$ red and $$2$$ black balls. Another box $$'B'$$ contains $$4$$ white, $$2$$ red and $$3$$ black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box $$'B'$$ is

A
$$\dfrac {7}{16}$$

B
$$\dfrac {9}{32}$$

C
$$\dfrac {7}{8}$$

D
$$\dfrac {9}{16}$$

Solution

For the bag $$A$$ we can see that there are $$2$$ white, $$3$$ red and $$2$$ black balls. Similarly, from bag $$B$$ we have $$4$$ white, $$2$$ red and $$3$$ black balls.

Probability of choosing a white and then a red ball from bag $$B$$ is given by $$=\dfrac{^4C_1 \times \ ^2C_1}{^9C_2}$$

Probability of choosing a white ball then a red ball from bag $$A$$ is given by $$=\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}$$

So, the probability of getting a white ball and then a red ball from bag $$B$$ is given by
$$\dfrac{\dfrac{^4C_1 \times \ ^2C_1 }{^9C_2}}{\dfrac{^4C_1 \times ^2C_1}{^9C_2}+\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}}$$ $$=\dfrac{\dfrac{2}{9}}{\dfrac{2}{7}+\dfrac{2}{9}} = \dfrac{2 \times 7}{18+14}=\dfrac{7}{16}$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 12

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Probability Test - 12

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Question 1

Both the statement are true

Both the statement are false

Statement - 1 is true, Statement - 2 is false

Statement - 1 is false, Statement - 2 is true.

Solution

Question 2

$$1/14$$

$$1/7$$

$$5/14$$

$$1/50$$

Solution

Question 3

$$\displaystyle \frac{3}{8}$$

$$\displaystyle \frac{1}{5}$$

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{2}{5}$$

Solution

Question 4

$$\displaystyle\frac{7}{32}$$

$$\displaystyle\frac{7}{16}$$

$$\displaystyle\frac{7}{64}$$

$$\displaystyle\frac{3}{16}$$

Solution

Question 5

$$P\left(\dfrac{C}{D}\right) =\mathrm{P}(\mathrm{C})$$

$$P\left(\dfrac{C}{D}\right) \geq \mathrm{P}(\mathrm{C})$$

$$P\left(\dfrac{C}{D}\right) <\mathrm{P}(\mathrm{C})$$

$$P\left(\dfrac{C}{D}\right) =\displaystyle \frac{\mathrm{P}(\mathrm{D})}{\mathrm{P}(\mathrm{C})}$$

Solution

Question 6

$$P(A)+P(\overline{B})$$

$$P(\overline{A})-P(\overline{B})$$

$$P(\overline{A})-P(B)$$

$$P(\overline{A})+P(\overline{B})$$

Solution

Question 7

$$\dfrac {5}{17}$$

$$\dfrac {11}{20}$$

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

$$\dfrac {8}{17}$$

Solution

Question 8

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{1}{3}$$

Solution

Question 9

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

$$\dfrac {3}{4}$$

$$\dfrac {3}{10}$$

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

Solution

Question 10

$$\dfrac {7}{16}$$

$$\dfrac {9}{32}$$

$$\dfrac {7}{8}$$

$$\dfrac {9}{16}$$

Solution

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