Probability Test 18

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Probability Test 18

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Question 1
1 / -0

A card is drawn from a well shuffled pack of 52 cards. Events A and B are defined as follows :
A : Getting a card of spade
B : Getting an ace, then A and B are .....

A
mutually exclusive and independent events

B
not mutually exclusive but independent events

C
mutually exclusive and not independent events

D
neither mutually exclusive nor independent events

Solution

A: Getting a card of spade

B : Getting an ace

i) Events A & B are not mutually exclusive

since the occurrence of one doesn't rule out the occurrence of others. For instance, getting an ace of spade

ii) Events A & B are independent as occurrence

of one is not dependent on other.

also $$P(A\cap B)=\dfrac {1}{52}$$ since ther is only $$1$$ ace of spades

& $$P(A)P(B)=\dfrac{13}{52}\times \dfrac{4}{52}= \dfrac{1}{52}$$

$$P(A\cap B)=P(A)P(B)\Rightarrow $$ Independent events

$$\Rightarrow $$ A & B are not mutually exclusive

but independent.
Question 2
1 / -0

If A and B are two events, such that P(A or B)=P(A), then

A
events A and B are mutually exclusive

B
events A and B are statistically independent

C
event B is a subset of event A

D
event A is a subset of event B

Solution

$$P\left( AorB \right) =P\left( A\cup B \right) =P\left( A \right) $$
In venn diagram it is shown that $$B$$ is a part of $$A$$ Hence probability of occurrance of event $$A$$ or event $$B$$ will be $$P(A)$$.
$$\therefore B$$ is subset of $$A$$
Question 3
1 / -0

Lot $$A$$ consists of $$3G$$ and $$2D$$ articles. Lot $$B$$ consists of $$4G$$ $$1D$$ article. A new lot $$C$$ is formed by taking $$3$$ articles from $$A$$ articles from $$A$$ and $$2$$ form $$B$$ . The probability that an article chosen at random from $$C$$ is

A
$$1 / 3$$

B
$$2 / 5$$

C
$$8 / 25$$

D
none

Solution

A] A : event it is chosen from lot A
$$ P(a) = \frac{3}{5}$$
$$ P(B) = \frac{2}{5}$$
P (Defective) $$= P(A)\times P(\frac{D}{A})+P(B)\times P(\frac{D}{B})$$
$$ = \frac{3}{5} \times \frac{2}{5}+\frac{2}{5}\times \frac{1}{5}$$
$$ = \frac{6}{25}+\frac{2}{25}$$
$$ = \frac{8}{25}$$
$$ \therefore P(D) = \frac{8}{25}$$
Question 4
1 / -0

A and B are two independent events such that $${ P\left( { A }^{ ' }\cap B \right) =\dfrac { 2 }{ 15 } }$$then P(B) is equal to :

A
$$\frac { 4 }{ 5 } $$

B
$$\frac { 1 }{ 6 } $$

C
$$\frac { 1 }{ 5 } $$

D
$$\frac { 5 }{ 6 } $$

Solution
Question 5
1 / -0

A bag contains $$6$$ red, $$4$$ white and $$8$$ blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.

A
$$\dfrac {2}{17}$$

B
$$\dfrac {3}{17}$$

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

D
$$\dfrac {4}{17}$$

Solution

$$E\rightarrow$$ Event of getting one is red, one is white and one $$6$$ red $$+4$$ white $$+8$$ blue balls $$=18$$ balls

Total outcomes $$=(\ ^{18}C_{3})$$

$$\underbrace { \bigodot }_{ R } \underbrace { \bigodot }_{ W } \underbrace { \bigodot }_{ B } \rightarrow$$ no. of fobourable element

$$=\ ^{6}C_{1}\times \ ^{4}C_{1}\times \ ^{8}C_{1}$$

$$=6\times 4\times 8$$

$$\therefore P(E)=\dfrac{6\times 4\times 8}{\ ^{18}C_{3}}=\dfrac{6\times 4\times 8\times 3\times 2\times 1}{18\times 17\times 16}$$

$$=\dfrac{4}{17}$$
Question 6
1 / -0

If A and B are such events that $$P(A)>0$$ and $$ P(B)\neq 1$$ then $$P\left(\dfrac{\bar{A}}{\bar{B}}\right)$$ is equal to-

A
$$1-P\left(\dfrac{A}{B}\right)$$

B
$$1-P\left(\dfrac{\bar{A}}{B}\right)$$

C
$$\dfrac{1-P(A\cup B)}{P(\bar{B})}$$

D
$$None$$

Solution
Question 7
1 / -0

If $$\dfrac {1+4p}{4},\dfrac {1+p}{4},\dfrac {1-2p}{2}$$ re probabilities of three mutually exclusive events, then

A
$$\dfrac {1}{3} \le \rho \le \dfrac {1}{2}$$

B
$$0 \le \rho \le \dfrac {2}{3}$$

C
$$\dfrac {1}{6} \le \rho \le \dfrac {1}{2}$$

D
$$0 \le \rho \le \dfrac {1}{2}$$

Solution
Question 8
1 / -0

Directions For Questions

$$8$$ players compete in a tournament, every one plays everyone else just once. The winner of a game gets $$1$$, the loser $$0$$ or each gets $$\dfrac{1}{2}$$ if the game is drawn. The final result is that every one gets a different score and the player playing placing second gets the same as the total of four bottom players.

...view full instructions

The score of the player placing II was

A
$$5\dfrac{1}{2}$$

B
$$6\dfrac{1}{2}$$

C
$$6$$

D
$$can't\ say$$

Solution

The total no.of matches held is given by $$\dfrac{8\times 7}{2}=28$$
So the winner is the one who has the score in between $$[28,14]$$
The second is the one who has less score than the first one.
Since it has many possibilities it cannot be determined.
Question 9
1 / -0

If $$\dfrac{1+4p}{p},\dfrac{1-p}{4}, \dfrac{1-2p}{2}$$ are propabilities of three mutually exclusive events, then-

A
$$\dfrac{-1}{5}\le p\le \dfrac{1}{2}$$

B
$$\dfrac{1}{3}\le p\le \dfrac{2}{3}$$

C
$$\dfrac{1}{6}\le p\le \dfrac{1}{2}$$

D
None of these

Solution
Question 10
1 / -0

A,B and C are three mutually exclusive and exhaustive events and $$P(B)=\dfrac{3}{2}P(A), P(C)=\dfrac{1}{3}P(B)$$ then the value of $$P(A)$$ is

A
1/3

B
1/2

C
1/6

D
none of these

Solution

Given, $$A, B$$ and $$C$$ are three mutually exclusive and exhaustive events and $$P(B)=\dfrac{3}{2}P(A), P(C)=\dfrac{1}{3}P(B)$$.

Since $$A, B, C$$ are mutually exclusive and exhaustive events then we've,

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

or, $$P(A)+\dfrac{3}{2}P(A)+\dfrac{1}{2}P(A)=1$$

or, $$3P(A)=1$$

or, $$P(A)=\dfrac{1}{3}$$.