Probability Test

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Probability Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

'X' speaks truth in $$60\%$$ of the cases and 'Y' in $$50\%$$ of the cases. The probability that they contradict each other while narrating the some incident, is

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

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

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

D
None of these

Solution

Let $$X$$ be the event that $$X$$ speaks the truth.

Let $$Y$$ be the event that $$Y$$ speaks the truth.

Then $$P(X)=\dfrac{60}{100}=\dfrac{3}{5}$$ and $$P(Y) =\dfrac{50}{100}=\dfrac{1}{2}$$

Also $$P(X-lie) =1-\dfrac{3}{5}=\dfrac{2}{5}$$ and $$P(Y-lie) = 1-\dfrac{1}{2}=\dfrac{1}{2}$$

Now
$$X$$ and $$Y$$ contradict each other $$=[X\:lies \:and\:Y\: true]\:or\:[X\:true\:and\:Y\:lies]$$

$$= P(X-lie) \times P(Y)+P(X)\times P(Y-lie)$$

$$=\dfrac{2}{5} \times \dfrac{1}{2}+\dfrac{3}{5} \times \dfrac{1}{2}$$

$$=\dfrac{2}{10} + \dfrac{3}{10}$$

$$=\dfrac{5}{10}$$

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

Therefore the probability that they contradict each other while narrating the some incident, is $$\dfrac{1}{2}$$
Question 2
1 / -0

If A and B are independent events of a random experiment such that $$P(A\cap B)=\dfrac{1}{6}$$ and $$P(\overline A\cap \overline B)=\dfrac{1}{3}$$, then $$P(A)$$ is equal to

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

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

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

D
None of these

Solution

Given $$P(A\cap B) =\cfrac{1}{6}$$ and $$P(\overline { A } \cap \overline { B } )=\cfrac{1}{3}$$
$$P(\overline { A } \cap \overline { B } )=P(\overline { A\cup B } )=1-P(A\cup B)$$
$$\Rightarrow P(A\cup B)=1-\cfrac{1}{3}=\cfrac{2}{3}$$
We know that $$P(A\cup B) =P(A)+P(B)-P(A\cap B)$$
$$\Rightarrow P(A)+P(B)=\cfrac{2}{3}+\cfrac{1}{6}=\cfrac{5}{6}$$
Given $$A$$ and $$B$$ are independent events , so we get $$P(A)P(B)=P(A\cap B)=\cfrac{1}{6}$$
By solving above equations , we get $$P(A)=\cfrac{1}{3}$$ and $$P(B)=\cfrac{1}{2}$$
Therefore option $$A$$ is correct.
Question 3
1 / -0

What is the probability of drawing a black card in a deck of cards?

A
$$\dfrac{5}{13}$$

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

C
$$\dfrac{5}{11}$$

D
None of these

Solution

Total number of cards in deck $$52$$.
Therefore, $$n(S)=52$$.
Let $$A$$ be the event of drawing a black card in a deck of cards.
$$\therefore n(A)=26$$.
Hence, the probability of drawing a black card in a deck of cards is $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{26}{52}=\dfrac{1}{2}$$.
Therefore, the probability of drawing a black card in a deck of cards is $$\dfrac{1}{2}$$.
Question 4
1 / -0

The probability that a leap year selected at random contains either $$53$$ Sundays or $$53$$ Mondays, is

A
$$\dfrac{7}{13}$$

B
$$\dfrac{2}{23}$$

C
$$\dfrac{3}{7}$$

D
None of these

Solution

Total number of days in a leap year is $$366$$.

It will contain $$52$$ weeks and $$2$$ days.

These two days can be
$$S=\{(Sun,Mon); (Mon,Tues); (Tues,Wed); (Wed, Thurs); (Thurs,Fri); (Fri,Sat); (Sat,Sun)\}$$

Therefore $$n(S)=7$$

Let $$A$$ be the event of getting $$53$$ sundays.

Therefore $$A=\{(Sun,Mon);(Sat,Sun)\}$$

For $$53$$ Sundays , probability is $$P(A)=\dfrac{2}{7}$$

Let $$B$$ be the event of getting $$53$$ mondays.

Therefore $$B=\{(Sun,Mon);(Mon,Tues)\}$$

For $$53$$ Mondays , probability is $$P(B)=\dfrac{2}{7}$$

This includes one ways where sunday and monday simultaneously Occur
(i.e) $$A\cap B= \{Sun, Mon\}$$

Probability for this is $$P(A\cap B)=\dfrac{1}{7}$$.

Hence required probability that a leap year selected at random contain $$53$$ sundays or $$53$$ mondays is
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$=\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{1}{7}=\dfrac{2+2-1}{7}$$

Therefore the required probability is $$\dfrac{3}{7}$$
Question 5
1 / -0

A box contains $$100$$ bulbs, out of which $$10$$ are defective. A sample of $$5$$ bulbs is drawn. The probability that none is defective is.

A
$$\left ( \dfrac{1}{10} \right )^5$$

B
$$\left ( \dfrac{1}{2} \right )^5$$

C
$$\left(\dfrac{9}{10}\right) $$

D
$$\left ( \dfrac{9}{10} \right )^5$$

Solution

The probability that a bulb is defective $$p=\dfrac{10}{100}=\dfrac{1}{10}$$.
The probability that a bulb is not defective $$q=1-p=\dfrac{9}{10}$$
Thus out of 5 bulbs, the probability that none of the bulbs are defective
$$=\dfrac{9}{10}\times \dfrac{9}{10}\times \dfrac{9}{10}\times \dfrac{9}{10}\times \dfrac{9}{10}$$

$$=(\dfrac{9}{10})^{5}$$
Question 6
1 / -0

If $$A$$ and $$B$$ are two independent events such that $$P(\overline A)= \dfrac{7}{10}$$, $$P(\overline B)= \alpha$$ and $$P(A \cup B)= \dfrac{8}{10}$$, then $$\alpha$$ is

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

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

C
$$\dfrac{6}{7}$$

D
None of these

Solution

We know $$P(A')=1-P(A)$$
$$\Rightarrow P(A)=1-\dfrac {7}{10}=\dfrac {3}{10}$$
So, $$P(A)=\dfrac{3}{10}$$
and $$P(B)=1-\alpha$$
Since they are both independent events,
So, $$P(A\cap{B})=P(A).P(B)$$
$$P(A\cap{B})=\dfrac{3}{10}(1-\alpha)$$
Now $$P(A\cup{B})=P(A)+P(B)-P(A\cap{B})$$
$$\Rightarrow \dfrac{8}{10}=\dfrac{3}{10}+1-\alpha-\dfrac{3}{10}(1-\alpha)$$
$$\Rightarrow 7\alpha=2$$
$$\Rightarrow \alpha=\dfrac{2}{7}$$
Question 7
1 / -0

Let $$A$$ and $$B$$ two events such that $$P(A\cap B) = \dfrac{1}{6}, P(A\cup B) = \dfrac{31}{45}$$ and $$P(\overline{B}) = \dfrac{7}{10}$$ then

A
$$A$$ and $$B$$ are independent

B
$$A$$ and $$B$$ are mutually exclusive

C
$$P\begin{pmatrix}\dfrac{A}{B}\end{pmatrix} < \dfrac{1}{6}$$

D
$$P\begin{pmatrix}\dfrac{B}{A}\end{pmatrix} < \dfrac{1}{6}$$

Solution

We have, $$P(A\cap B)=\dfrac{1}{6}, P(A\cup B)=\dfrac{31}{45}$$ and $$P(\overline{B})=\dfrac{7}{10}$$

$$\therefore P(B)=1-\dfrac{7}{10}=\dfrac{3}{10}$$

Now using $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$\Rightarrow \dfrac{31}{45}=P(A)+\dfrac{3}{10}-\dfrac{1}{6}=P(A)+\dfrac{2}{15}$$
$$\Rightarrow P(A)=\dfrac{31}{45}-\dfrac{2}{15}=\dfrac{25}{45}=\dfrac{5}{9}$$

Check, $$P(A)P(B)=\dfrac{5}{9}\cdot \dfrac{3}{10}=\dfrac{1}{6}=P(A\cap B)$$

Hence both the events $$A$$ and $$B$$ are independent to each other
Question 8
1 / -0

Which of the following is true for two independent events?

A
Multiplication rule is not valid

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

C
Both are correct

D
None is correct

Solution

We know that when $$A$$ and $$B$$ are independent events, then the intersection of two events is given by $$P(A\cap B)=P(A)\;P(B)$$.

Hence option $$(B)$$ is correct.
Question 9
1 / -0

A policeman fires four bullets at a dacoit. The probability that the dacoit will be killed by one bullet is 0.6. What is the probability that the dacoit is still alive?

A
$$0.064$$

B
$$0.0256$$

C
$$0.32$$

D
$$0.16$$

Solution

The probability of death for each bullet is $$0.6$$.

That means the probability for staying alive is $$1-0.6=0.4$$.

Given, the policeman fires $$4$$ bullets at a dacoit.

So to stay alive the dacoit must miss four bullets.

Hence the probability that the dacoit is still alive is $$(0.4) \times (0.4) \times (0.4) \times (0.4)=(0.4)^4=0.0256$$
Question 10
1 / -0

The probabilities that three men hit a target are 1/6, 1/4 and 1/3. Each man shoots once at the target. What is the probability that exactly one of them hits the target?

A
11/72

B
21/72

C
31/72

D
3/4

Solution

Let the probability of A hitting the target $$=\dfrac{1}{6}$$
Let the probability of B hitting the target $$=\dfrac{1}{4}$$
Let the probability of C hitting the target $$=\dfrac{1}{3}$$
The probability that exactly one of them hits the target (it means only one will hit the target while others two cannot)
$$\cfrac { 1 }{ 6 } \times \cfrac { 3 }{ 4 } \times \cfrac { 2 }{ 3 } +\cfrac { 5 }{ 6 } \times \cfrac { 3 }{ 4 } \times \cfrac { 1 }{ 3 } +\cfrac { 5 }{ 6 } \times \cfrac { 1 }{ 4 } \times \cfrac { 2 }{ 3 } $$
$$=\dfrac{31}{72}$$

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Probability Test - 42

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Probability Test - 42

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Question 1

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

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

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

None of these

Solution

Question 2

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

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

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

None of these

Solution

Question 3

$$\dfrac{5}{13}$$

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

$$\dfrac{5}{11}$$

None of these

Solution

Question 4

$$\dfrac{7}{13}$$

$$\dfrac{2}{23}$$

$$\dfrac{3}{7}$$

None of these

Solution

Question 5

$$\left ( \dfrac{1}{10} \right )^5$$

$$\left ( \dfrac{1}{2} \right )^5$$

$$\left(\dfrac{9}{10}\right) $$

$$\left ( \dfrac{9}{10} \right )^5$$

Solution

Question 6

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

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

$$\dfrac{6}{7}$$

None of these

Solution

Question 7

$$A$$ and $$B$$ are independent

$$A$$ and $$B$$ are mutually exclusive

$$P\begin{pmatrix}\dfrac{A}{B}\end{pmatrix} < \dfrac{1}{6}$$

$$P\begin{pmatrix}\dfrac{B}{A}\end{pmatrix} < \dfrac{1}{6}$$

Solution

Question 8

Multiplication rule is not valid

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

Both are correct

None is correct

Solution

Question 9

$$0.064$$

$$0.0256$$

$$0.32$$

$$0.16$$

Solution

Question 10

11/72

21/72

31/72

3/4

Solution

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