Probability Test 20

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Probability Test 20

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Question 1
1 / -0

Let A and B be two events such that P(A) $$= \frac{3}{8}$$, P(B) $$= \frac{5}{8}$$ and P(A $$\cup$$ B) $$= \frac{3}{4}$$. Then $$P(A|B). P(A'|B)$$ is equal to

A
$$\frac{2}{5}$$

B
$$\frac{3}{8}$$

C
$$\frac{3}{10}$$

D
$$\frac{6}{25}$$

Solution

Here, P(A) $$= \displaystyle \frac{3}{8}$$, P(B) $$=\displaystyle \frac{5}{8}$$ and P(A $$\cup$$ B) $$= \displaystyle \frac{3}{4}$$

$$\because P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$\Rightarrow P(A \cap B) = \displaystyle \frac{3}{8} + \frac{5}{8} - \frac{3}{4} = \frac{3 + 5 - 6}{8} = \frac{2}{8} = \frac{1}{4}$$
$$\because P(A / B) = \displaystyle \frac{P(A \cap B)}{P(B)} = \frac{1/4}{5/8} = \frac{8}{20} = \frac{2}{5}$$

And $$P(A' / B) = \displaystyle \frac{P(A' \cap B)}{P(B)} = \frac{P(B) - P(A \cap B)}{P(B)}$$
$$=\displaystyle \frac{\frac{5}{8} - \frac{1}{4}}{\frac{5}{8}} = \frac{\frac{5 - 2}{8}}{\frac{5}{8}} = \frac{2}{5}$$
$$\therefore P(A/B) \cdot P(A'/B) =\displaystyle \frac{2}{5} . \frac{3}{5} = \frac{6}{25}$$
Question 2
1 / -0

If two events are independent, then

A
they must be mutually exclusive

B
the sum of their probabilities must be equal to 1

C
(A) and (B) are both correct

D
None of the above is correct

Solution

If two events A and B are independent, then we know that
$$P(A \cap B) = P(A) \cdot P(B), P(A) \neq 0, P(B) \neq 0$$
Since, A and B have a common outcome.
Further, mutually exclusive events never have a common outcome.
In other words, two independents events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely, i.e., two mutually exclusive events having non-zero probabilities of outcome cannot be independent.
Question 3
1 / -0

A card is drawn from a well shuffled deck of 52 playing cards. The probability that the card drawn not an ace of diamond is E. Then number of favourable outcomes of E is:

A
$$4$$

B
$$13$$

C
$$48$$

D
$$51$$

Solution

The number of playing cards in a deck = 52.
One card is drawn from this well shuffled deck.
The probability that the card drawn is an ace of diamond is E.

∴ The number of favourable outcomes of E = 52 – 1 = 51
Question 4
1 / -0

Two events $$A$$ and $$B$$ are said to be mutually independent, if:

A
$$P(A)=P(B)$$

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

C
$$P\left( \overline { A } \overline { B } \right) =\left[ 1-P(A) \right] \left[ 1-P(B) \right] $$

D
$$A$$ and $$B$$ are mutually disjoint

Solution

$$P(A)=1-P(\overline { A } );P(B)=1-P(\overline { B } )$$
$$P(\overline { AB } )=P(\overline { A } )\times P(\overline { B } )=\left[ 1-P( { A } ) \right] \left[ 1-P( { B } ) \right] $$
Question 5
1 / -0

5 cards are drawn at random from a well shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that there are exactly 3 hearts is

A
$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}+^{13}c_{4}+^{13}c_{5}}$$

B
$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

C
$$\displaystyle \frac{^{13}c_{3}\times ^{39}c_{2}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

D
$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{13}c_{4}\times ^{13}c_{5}}$$

Solution

The cases possible are:
(i) 3 H, 2 non-H: $$_{ 3 }^{ 13 }{ C }*_{ 2 }^{ 39 }{ C }$$
(ii) 4 H, 1 non-H: $$_{ 4 }^{ 13 }{ C }*_{ 1 }^{ 39 }{ C }$$
(iii) 5H: $$_{ 5 }^{ 13 }{ C }$$
Hence, probability = $$\dfrac { _{ 3 }^{ 13 }{ C }*_{ 2 }^{ 39 }{ C } }{ _{ 3 }^{ 13 }{ C }*_{ 2 }^{ 39 }{ C }+_{ 4 }^{ 13 }{ C }*_{ 1 }^{ 39 }{ C }+_{ 5 }^{ 13 }{ C } } $$
Hence, (c) is correct.
Question 6
1 / -0

A bag contains 6 balls. Two balls are drawn and found to be red. The probability that five balls in the bag are red

A
$$\displaystyle \frac{5}{6}$$

B
$$\displaystyle \frac{12}{17}$$

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

D
$$\displaystyle \frac{2}{7}$$

Solution

Let A be the event that we draw two red balls.
Let B be the event that five balls are red.
We need to find $$P(B|A)$$.
Required probability = $$P(\dfrac { B }{ A } )=\dfrac { P(\dfrac { A }{ B } )\times P(B) }{ P(A) } $$

Now, $$P(A)=P(\dfrac { A }{ 2R } )\times P(2R)+P(\dfrac { A }{ 3R } )\times P(3R)+P(\dfrac { A }{ 4R } )\times P(4R)+P(\dfrac { A }{ 5R } )\times P(5R)+P(\dfrac { A }{ 6R } )\times P(6R)$$
There can be $$5$$ cases: 6 Red or 5 Red or 4 Red or 3 Red or 2 Red. We assume that each case has a probability of $$\dfrac { 1 }{ 5 } $$.

Thus, $$P(A)=\dfrac { _{ 2 }^{ 2 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 3 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 4 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 5 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 6 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } =\dfrac { 7 }{ 15 } $$

Again, $$P(\dfrac { A }{ B } )=\dfrac { _{ 2 }^{ 5 }{ C } }{ _{ 2 }^{ 6 }{ C } } =\dfrac { 2 }{ 3 } $$
Therefore, required probability $$=$$ $$\dfrac { \dfrac { 2 }{ 3 } \times \dfrac { 1 }{ 5 } }{ \dfrac { 7 }{ 15 } } =\dfrac { 2 }{ 7 } $$
Question 7
1 / -0

A letter is known to have come form TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. The probability that the letter has come from CALCUTTA is

A
$$\displaystyle \frac{4}{11}$$

B
$$\displaystyle \frac{7}{11}$$

C
$$\displaystyle \frac{1}{22}$$

D
$$\displaystyle \frac{21}{22}$$

Solution

Let's calculate the total probability of getting $$TA$$
$$P\left( TA \right) =P\left( TA/TATANAGAR \right) \times P\left( TATANAGAR \right) +P\left( TA/CALCUTTA \right) \times P\left( CALCUTTA \right) \\ TATANAGAR\quad are\quad TA,AT,TA,AN,NA,AG,GA,AR$$
Now $$\quad P\left( TA/TATANAGAR \right) ={ \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) }/{ \left( \begin{matrix} 8 \\ 1 \end{matrix} \right) }={ 2 }/{ 8 }$$
and $$ P\left( TA/CALCUTTA \right) ={ 1 }/{ \left( \begin{matrix} 7 \\ 1 \end{matrix} \right) }={ 1 }/{ 7}$$
since both CALCUTTA and TATANAGAR are equally likely
$$ \Rightarrow P\left( TA \right) ={ 1 }/{ 2 }\times \left\{ { 2 }/{ 8 }+{ 1 }/{ 7 } \right\} ={ 11 }/{ 56 }$$
Hence by Bayes' theorem :
$$P\left( CALCUTTA/TA \right) =\dfrac { P\left( TA/CALCUTTA \right) P\left( CALCUTTA \right) }{ P\left( TA \right) } =\dfrac { 1 }{ 7 } \times \dfrac { 1 }{ 2 } \times \dfrac { 56 }{ 11 } =\dfrac { 4 }{ 11 }$$
Question 8
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A bag contains 4 white and 5 black balls. A second bag contains 3 (identical) white and 6 (identical) black balls. One bag is chosen at random and a ball is drawn. Its colour is noted and the ball replaced. This is repeated four times. It was found that of these four, one was white and 3 were black. The $$n$$ the probability that the first bag was chosen is

A
$$\dfrac{162}{287}$$

B
$$\dfrac{125}{287}$$

C
$$\dfrac{250}{280}$$

D
$$\dfrac{125}{574}$$

Solution

If $$A$$ is the event of choosing the first bag and $$B$$ the event of choosing the second bag. $$P(A) = P(B) = 12 $$
Let $$C$$ be the event of one white and $$3$$ black being drawn .
Using Baye's theorem
$$P(A|C)=\dfrac { P(C|A)*P(A) }{ P(C|A)*P(A)+P(C|B)*P(B) } =\dfrac { 1/2\times (4/9)^{ 2 }\times (5/9)^{ 3 } }{ 1/2\times (4/9)^{ 2 }\times (5/9)^{ 3 }+1/2\times (1/3)^{ 2 }\times (2/3)^{ 3 } } =125/287$$
Question 9
1 / -0

5 cards are drawn at random from a well shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that there are 4 hearts is

A
$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}+^{13}C_{4}+^{13}C_{5}}$$

B
$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{39}C_{2}+^{13}c_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

C
$$\displaystyle \frac{^{13}C_{4}\times ^{39}C_{1}}{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

D
$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{13}C_{4}\times ^{13}C_{5}}$$

Solution

The cases possible are:
(i) 3 H, 2 non-H: $$^{ 13 }{ C_3 }*^{ 39 }{ C_2 }$$
(ii) 4 H, 1 non-H: $$^{ 13 }{ C_4 }*^{ 39 }{ C_1 }$$
(iii) 5H: $$^{ 13 }{ C_5 }$$
Hence, probability = $$\dfrac { ^{ 13 }{ C_4 }*^{ 39 }{ C_1 } }{^{ 13 }{ C_3 }*^{ 39 }{ C_2 }+^{ 13 }{ C_4 }*^{ 39 }{ C_1 }+^{ 13 }{ C_5 } } $$
Question 10
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$$A$$ and $$B$$ are two independent witnesses in a case. The probability that $$A$$ will speak truth is $$x$$ and the probability that $$B$$ will speak the truth is $$y$$. $$A$$ and $$B$$ agree in a certain statement. The probability that the statement is true is

A
$$\displaystyle \frac{xy}{1-xy}$$

B
$$\displaystyle \frac{xy}{1-x-y+2xy}$$

C
$$\displaystyle \frac{2{x}y}{x-y}$$

D
$$\displaystyle \frac{xy}{1-x-y}$$

Solution

Let $$K$$ be the event that both of them agree, $$L$$ be the event that they both speak the truth and $$M$$ be the event that they both lie.
Probability that both $$A$$ and $$B$$ speak the truth = $$P(L)=xy$$
Probability that both $$A$$ and $$B$$ lie = $$P(M)=(1-x)(1-y)=1-x-y+xy$$
We need to find $$P\left(\dfrac { L }{ K } \right)$$
$$P\left(\dfrac { L }{ K } \right)=\dfrac { P\left(L\right)P\left(\dfrac { K }{ L } \right) }{ P\left(L\right)P\left(\dfrac { K }{ L } \right)+P\left(M\right)P\left(\dfrac { K }{ M } \right) }=\dfrac { xy\times 1 }{ \left(xy\right)1+\left(1-x-y+xy\right)1 }=\dfrac { xy }{ 1-x-y+2xy } $$

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Probability Test 20

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Probability Test 20

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SHARING IS CARING

Question 1

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

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

$$\frac{3}{10}$$

$$\frac{6}{25}$$

Solution

Question 2

they must be mutually exclusive

the sum of their probabilities must be equal to 1

(A) and (B) are both correct

None of the above is correct

Solution

Question 3

$$4$$

$$13$$

$$48$$

$$51$$

Solution

Question 4

$$P(A)=P(B)$$

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

$$P\left( \overline { A } \overline { B } \right) =\left[ 1-P(A) \right] \left[ 1-P(B) \right] $$

$$A$$ and $$B$$ are mutually disjoint

Solution

Question 5

$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}+^{13}c_{4}+^{13}c_{5}}$$

$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

$$\displaystyle \frac{^{13}c_{3}\times ^{39}c_{2}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{13}c_{4}\times ^{13}c_{5}}$$

Solution

Question 6

$$\displaystyle \frac{5}{6}$$

$$\displaystyle \frac{12}{17}$$

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

$$\displaystyle \frac{2}{7}$$

Solution

Question 7

$$\displaystyle \frac{4}{11}$$

$$\displaystyle \frac{7}{11}$$

$$\displaystyle \frac{1}{22}$$

$$\displaystyle \frac{21}{22}$$

Solution

Question 8

$$\dfrac{162}{287}$$

$$\dfrac{125}{287}$$

$$\dfrac{250}{280}$$

$$\dfrac{125}{574}$$

Solution

Question 9

$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}+^{13}C_{4}+^{13}C_{5}}$$

$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{39}C_{2}+^{13}c_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

$$\displaystyle \frac{^{13}C_{4}\times ^{39}C_{1}}{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{13}C_{4}\times ^{13}C_{5}}$$

Solution

Question 10

$$\displaystyle \frac{xy}{1-xy}$$

$$\displaystyle \frac{xy}{1-x-y+2xy}$$

$$\displaystyle \frac{2{x}y}{x-y}$$

$$\displaystyle \frac{xy}{1-x-y}$$

Solution

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