Probability Test

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Probability Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

If A & B are independent events such that $$P(B) = \dfrac{2}{7}, P(A\cup \overline{B})=0.8$$, then $$P(A)$$ is equal to

A
$$0.1$$

B
$$0.2$$

C
$$0.3$$

D
$$0.4$$

Solution

$$P(A\cup \overline{B}) =0.8$$
$$P(B) = \cfrac{2}{7}$$
$$\Rightarrow P(A)+P(\overline{B})-P(A\cap \overline{B})=0.8$$ & $$P(\overline{B})=1-\cfrac{2}{7}$$
$$\Rightarrow P(A)+P(\overline{B})-P(A)P(\overline{B})=0.8$$ & $$P(\overline{B})=\cfrac{5}{7}$$

$$\Rightarrow P(A)+\cfrac{5}{7}-\cfrac{5}{7}P(A)=0.8\Rightarrow \cfrac{2}{7}P(A) = \cfrac{3}{35}$$

$$\Rightarrow P(A) = 0.3$$
Question 2
1 / -0

If $$P(A) = \cfrac{1}{4}, P(\overline{B})=\cfrac{1}{2}$$ and $$P(A\cup B) = \cfrac{5}{9}$$, then $$P(A/B)$$ is

A
$$\cfrac{7}{36}$$

B
$$\cfrac{7}{9}$$

C
$$\cfrac{7}{18}$$

D
$$\cfrac{7}{72}$$

Solution

We have,
$$P(A\cup B) =\cfrac{5}{9}$$
$$\Rightarrow P(A)+P(B)-P(A\cap B)=\cfrac{5}{9}$$
$$\Rightarrow \cfrac{1}{4}+\cfrac{1}{2}-P(A\cap B) = \cfrac{5}{9}$$
$$\Rightarrow P(A\cap B)=\cfrac{7}{36}$$
$$\Rightarrow P(B)P(A/B)=\cfrac{7}{36}$$
$$\Rightarrow \cfrac{1}{2}\times P(A/B)=\cfrac{7}{36}\Rightarrow P(A/B) =\cfrac{7}{18}$$
Question 3
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For 3 events A, B and C, P(exactly one of the events A or B occurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p & P (all the three events occurs simultaneously) = $$p^2$$. Where $$ o < p < \frac{1}{2}$$. then the probability of at lest one of three events, A, B & C occurring is

A
$$\dfrac{3p+2p^2}{2}$$

B
$$\dfrac{p+2p^2}{2}$$

C
$$\dfrac{3p+p^2}{2}$$

D
$$\dfrac{3p+2p^2}{4}$$

Solution

$$P(A) -P(B) -2P(A\cap B) = p$$ ...(i)
$$P(B) + P(C) -2P(B\cap C) = p$$ ...(ii)
$$P(C) + P(A) -2P(C\cap A) = p$$ ...(iii)
and, $$P(A \cap B\cap C) = p^2$$ ...(iv)
Adding (i), (ii) and (iii), we get
$$2[P(A)+P(B)+P(C)+P(A\cap B)-P(B\cap C)-P(A\cap C)]=3p$$
$$\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C) = \cfrac{3p}2$$
$$\therefore $$ Required probability
$$=P(A\cup B \cup C)$$
$$=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$$
$$=\cfrac{3p}{2}+p^2=\cfrac{3p+2p^2}{2}$$
Question 4
1 / -0

'X' speaks truth in $$60%$$ & 'y' is $$50%$$ of the cases. The probability that they contradict each other while narrating the same incident, is

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

B
$$\dfrac{1}{8}$$

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

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

Solution

Given:
Probability of $$X$$ speaking truth, $$P(X)= 60\%=0.6$$
Probability of $$Y$$ speaking truth, $$P(Y) = 50\%=0.5$$
To find:
Probability of contradiction.
$$P(X')=0.4, P(Y')=0.5$$
Contradiction occurs when X speaks truth and Y speaks false or X speaks false and Y speaks truth. And as all the events are independent,
i.e.,, Required probability = $$P(X\cap Y')+P(X'\cap Y)=P(X)\times P(Y')+P(X')\times P(Y)$$
$$\implies = 0.6\times 0.5+0.4\times 0.5=0.3+0.2=0.5=\dfrac 12$$
Question 5
1 / -0

An integer is chosen at random from first 200 positive integers. The probability that the integer chosen is divisible by 6 or 8 is

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

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

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

D
None of these

Solution

One integer can be chosen out of $$200$$ integer in $$^{200}C_1$$ ways.
Let A be event than an integer selected is divisible by 6 and B that it is divisible by 8.
Then, $$P(A) = \dfrac{33}{200}, P(B)=\dfrac{25}{200}$$ and $$P(A\cap B) = \dfrac{8}{192}$$
$$\therefore$$ Required probability $$=P(A \cup B)$$
$$=P(A)+P(B)-P(A\cap B)$$$$=\cfrac{33}{200}+\cfrac{25}{200}-\cfrac{8}{192}=\cfrac{1}{4}$$
Question 6
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The probability of simultaneous occurrence of 2 events A & B is $$p$$. If the probability that exactly one of A, B occurs is $$q$$, then which of the following alternatives is INCORRECT?

A
$$P(\overline{A})+P(\overline{B})= 2+2q-p$$

B
$$P(\overline{A})+P(\overline{B})= 2+2p-q$$

C
$$P((A\cap B) / (A \cup B)) = \dfrac{P}{p+q}$$

D
$$P(\overline{A}\cap \overline{B})= 21-p-q$$

Solution

$$P(A\cap B) =p$$ and $$P(A) +P(B)-2P(A\cap B)=q$$
$$\Rightarrow P(A)+P(B)-2p=q$$
$$\Rightarrow P(A)+P(B)=2p-q$$
$$\Rightarrow 1-P(\overline{A})+1-P(\overline{B})=2p+q$$
$$\Rightarrow P(\overline{A})+P(\overline{B})=2-2p-q$$
So, alternative (b) is correct.
Now,
$$P\{(A\cap B)/(A\cup B)\} = \cfrac{P[(A\cap B)\cap (A\cup B)]}{P(A\cup B)}$$
$$\Rightarrow P\{(A\cap B)/(A\cup B)\} = \cfrac{P(A\cap B)}{P(A\cup B)}$$
$$\Rightarrow P\{(A\cap B)/(A\cup B)\} = \cfrac{P[(A\cap B)}{P(A)+P(B)-P(A\cap B)}$$
$$\Rightarrow P\{(A\cap B)/(A\cup B)\} = \cfrac{P}{2p+q-p} = \cfrac{p}{p+q}$$
So, alternative (c) is correct.
Finally,
$$P(\overline{A}\cap \overline{B}) = P(\overline{A\cup B}) = 1-P(A\cup B)$$
$$\Rightarrow P(\overline{A}\cap \overline{B}) = 1-[P(A)+P(B)-P(A\cap B)]$$
$$\Rightarrow P(\overline{A}\cap \overline{B}) = 1-[2p+q-p]=1-p-q$$
So, alternative (d) is correct.
Hence, alternative (a) is incorrect.
Question 7
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If $$A$$ and $$B$$ are two event such that $$P(A)=\cfrac{3}{4}$$ and $$P(B)=\cfrac{5}{8}$$, then

A
$$P(A\cup B) \ge \dfrac{3}{4}$$

B
$$P(\bar{A}\cap B)\le \dfrac{1}{4}$$

C
$$\dfrac{3}{8}\le P({A}\cap B) \le \dfrac{5}{8}$$

D
All of these

Solution

$$P\left( A\cup B \right) \ge max\left\{ P(A),P(B) \right\} $$

Here $$P(A)=\dfrac { 3 }{ 4 } ,P(B)=\dfrac { 5 }{ 8 } $$

$$\Rightarrow P\left( A\cup B \right) \ge \dfrac { 3 }{ 4 } $$

So, option A is correct.

$$P\left( A\cap B \right) \le min\left\{ P(A),P(B) \right\} \\ \Rightarrow P\left( A\cap B \right) \le \dfrac { 5 }{ 8 } $$ ......(i)
We know $$ P\left( A\cup B \right) =P(A)+P(B)-P\left( A\cap B \right) $$
$$\Rightarrow P\left( A\cap B \right) =P(A)+P(B)-P\left( A\cup B \right) \\ \Rightarrow P\left( A\cap B \right) \ge P(A)+P(B)-1\\ \Rightarrow P\left( A\cap B \right) \ge \dfrac { 3 }{ 4 } +\dfrac { 5 }{ 8 } -1\\ \Rightarrow P\left( A\cap B \right) \ge \dfrac { 3 }{ 8 }$$ .......(ii)

From (i) and (ii), we have

$$\dfrac { 3 }{ 8 } \le P\left( A\cap B \right) \le \dfrac { 5 }{ 8 } $$

So, option C is correct.

$$\dfrac { 3 }{ 8 } \le P\left( A\cap B \right) \le \dfrac { 5 }{ 8 } \\ \Rightarrow -\dfrac { 5 }{ 8 } \le -P\left( A\cap B \right) \le -\dfrac { 3 }{ 8 } $$
$$\Rightarrow P(B)-\dfrac { 5 }{ 8 } \le P(B)-P\left( A\cap B \right) \le P(B)-\dfrac { 3 }{ 8 } \\ \Rightarrow \dfrac { 5 }{ 8 } -\dfrac { 5 }{ 8 } \le P\left( \overset { \_ }{ A } \cap B \right) \le \dfrac { 5 }{ 8 } -\dfrac { 3 }{ 8 } \\ \Rightarrow 0\le P\left( \overset { \_ }{ A } \cap B \right) \le \dfrac { 1 }{ 4 } \\ \Rightarrow P\left( \overset { \_ }{ A } \cap B \right) \le \dfrac { 1 }{ 4 } $$
Question 8
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For any 2 events A and B in a sample space which one of the following is incorrect?

A
$$P(A/B)\ge \dfrac{P(A)+P(B)-1}{P(B)}, P(B) \neq 0$$, is always true

B
$$P(A\cap \overline{B})=P(A)-P(A\cap B)$$ doen't hold

C
$$P(A\cup B)=1-P(\overline{A})P(\overline{B})$$ if A & B are independent

D
$$P(A\cup B)=1-P(\overline{A})P(\overline{B})$$, if A & B are disjoint.

Solution

Consider $$2$$ events $$A$$ $$\xi$$ $$B$$
$$P\left( \dfrac { A }{ B } \right)\ge\dfrac{P(A)+P(B)-1}{P(B)},P(B)\neq 0,$$ is always true.
This condition is not correct.
$$P\left( \dfrac { A }{ B } \right)=\dfrac{Number\;of\;elementary\;events\;favorable\;A\cap B}{Number\;of\;elementary\;events\;favorable\;to\;B}$$

$$P\left( \dfrac { A }{ B } \right)=\dfrac{n(A\cap B)}{n(B)}=\dfrac{\dfrac{n(A\cap B)}{n(s)}}{\dfrac{n(B)}{n(s)}}$$

$$\therefore P\left( \dfrac { A }{ B } \right)=\dfrac{(A\cap B)}{(B)}$$
Similarly $$P\left( \dfrac { B }{ A } \right)=\dfrac{(A\cap B)}{(A)}$$
Hence, the answer is $$P\left( \dfrac { A }{ B } \right)\ge\dfrac{P(A)+P(B)-1}{P(B)},P(B)\neq 0,$$ is always true.
Question 9
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A coin is tossed and a six-sided die is rolled. Find the probability of getting a head on the coin and a $$6$$ on the die.

A
$$\dfrac{1}{16}$$

B
$$\dfrac{1}{12}$$

C
$$\dfrac{1}{18}$$

D
None of these

Solution

A coin is tossed and a six-sided die is rolled.

We have to find the probability of getting head on the coin and a $$6$$ on the die.

Probability of getting head in the coin is $$\dfrac{1}{2}$$.

Probability of getting $$6$$ on the die is $$\dfrac{1}{6}$$.

Probability of getting head on the coin and a $$6$$ on the die is $$\dfrac{1}{2} \times \dfrac{1}{6}= \dfrac{1}{12}$$
Question 10
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If $$A$$ and $$B$$ are arbitrary events, then

A
$$P(A\cap B)\geq P(A)+P(B)$$

B
$$P(A\cup B)\leq P(A)+P(B)$$

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

D
None of these

Solution

If $$A$$ and $$B$$ are arbitrary events, we know that
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
SInce $$P(A\cup B)$$ can be written as $$P(A)+P(B)$$ $$-$$ some positive quantity, we can say that
$$P(A\cup B)\leq P(A)+P(B)$$..

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Re-Attempt Weekly Quiz Competition

Probability Test - 40

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Probability Test - 40

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SHARING IS CARING

Question 1

$$0.1$$

$$0.2$$

$$0.3$$

$$0.4$$

Solution

Question 2

$$\cfrac{7}{36}$$

$$\cfrac{7}{9}$$

$$\cfrac{7}{18}$$

$$\cfrac{7}{72}$$

Solution

Question 3

$$\dfrac{3p+2p^2}{2}$$

$$\dfrac{p+2p^2}{2}$$

$$\dfrac{3p+p^2}{2}$$

$$\dfrac{3p+2p^2}{4}$$

Solution

Question 4

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

$$\dfrac{1}{8}$$

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

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

Solution

Question 5

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

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

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

None of these

Solution

Question 6

$$P(\overline{A})+P(\overline{B})= 2+2q-p$$

$$P(\overline{A})+P(\overline{B})= 2+2p-q$$

$$P((A\cap B) / (A \cup B)) = \dfrac{P}{p+q}$$

$$P(\overline{A}\cap \overline{B})= 21-p-q$$

Solution

Question 7

$$P(A\cup B) \ge \dfrac{3}{4}$$

$$P(\bar{A}\cap B)\le \dfrac{1}{4}$$

$$\dfrac{3}{8}\le P({A}\cap B) \le \dfrac{5}{8}$$

All of these

Solution

Question 8

$$P(A/B)\ge \dfrac{P(A)+P(B)-1}{P(B)}, P(B) \neq 0$$, is always true

$$P(A\cap \overline{B})=P(A)-P(A\cap B)$$ doen't hold

$$P(A\cup B)=1-P(\overline{A})P(\overline{B})$$ if A & B are independent

$$P(A\cup B)=1-P(\overline{A})P(\overline{B})$$, if A & B are disjoint.

Solution

Question 9

$$\dfrac{1}{16}$$

$$\dfrac{1}{12}$$

$$\dfrac{1}{18}$$

None of these

Solution

Question 10

$$P(A\cap B)\geq P(A)+P(B)$$

$$P(A\cup B)\leq P(A)+P(B)$$

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

None of these

Solution

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