Probability Test

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Probability Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

For any two independent events $$E_{1}$$ and $$E_{2}$$ in a space $$S, P[(E_{1} \cup E_{2})\cap (E_{1}\cap E_{2})]$$ is equal to

A
$$\leq \dfrac {1}{4}$$

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

C
$$\geq \dfrac {1}{2}$$

D
$$> \dfrac {1}{2}$$

Solution

$$P(E)=P[\left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( { E }_{ 1 }\cap { E }_{ 2 } \right) ]$$

Using De Morgans law

$$\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =\left( { E }_{ 1 }^{ \prime }\cup { E }_{ 2 }^{ \prime } \right) ={ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) }^{ \prime }$$

$$ \Rightarrow P(E)=P[\left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap { \left( { E }_{ 1 }\cup { E }_{ 2 } \right) }^{ \prime }]$$

Now $$\left( A\cap A\prime \right) $$ is always $$\phi$$ set

$$\Rightarrow P(E)=\phi =0$$

Out of the options

$$ \Rightarrow P(E)\le \dfrac { 1 }{ 4 } $$

So option $$A$$ is correct.
Question 2
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If $$A$$ and $$B$$ are two events, then, $$1+P\left( A\cap B \right) -P\left( B \right) -P\left( A \right) $$ is equal to

A
$$P\left( \overline { A } \cup \overline { B } \right) $$

B
$$P\left( A\cap \overline { B } \right) $$

C
$$P\left( \overline { A } \cap B \right) $$

D
$$P\left( A\cup B \right) $$

E
$$P\left( \overline { A } \cap \overline { B } \right) $$

Solution

We know that for two events,
$$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow P\left( A \right) +P\left( B \right) =P\left( A\cup B \right) +P\left( A\cap B \right) $$
$$1+P\left( A\cap B \right) -P\left( B \right) -P\left( A \right) $$
$$=1+P\left( A\cap B \right) -\left( P\left( B \right) +P\left( A \right) \right)$$
$$=1+P\left( A\cap B \right) -\left( P\left( A\cup B \right) +P\left( A\cap B \right) \right) $$
$$=1+P\left( A\cap B \right) -P\left( A\cup B \right) -P\left( A\cap B \right) $$
$$=1-P\left( A\cup B \right) =P\overline { \left( A\cup B \right) } =P\left( \overline { A } \cap \overline { B } \right) $$
Question 3
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If the events $$A$$ and $$B$$ are independent and if $$P(\overline {A}) = \dfrac {2}{3}, P(\overline {B}) = \dfrac {2}{7}$$, then $$P(A\cap B)$$ is equal to

A
$$\dfrac {4}{21}$$

B
$$\dfrac {3}{21}$$

C
$$\dfrac {5}{21}$$

D
$$\dfrac {2}{21}$$

E
$$\dfrac {1}{21}$$

Solution

Given, $$P(\overline {A}) = \dfrac {2}{3}$$

Therefore, $$P(A) = 1 - P(\overline {A}) = 1- \dfrac {2}{3} = \dfrac {1}{3}$$

and $$P(\overline {B}) = \dfrac {2}{7}$$

$$\therefore P(B) = 1 - P(\overline {B}) = 1 -\dfrac {2}{7} = \dfrac {5}{7}$$

Given, $$A$$ and $$B$$ are independent events.

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

$$= \dfrac {1}{3}\cdot \dfrac {5}{7} = \dfrac {5}{21}$$
Question 4
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The probabilities that Mr. $$A$$ and Mr. $$B$$ will die within a year are $$\dfrac { 1 }{ 2 } $$ and $$\dfrac { 1 }{ 3 } $$ respectively, then the probability that only one of them will be alive at the end of the year, is

A
$$\dfrac { 5 }{ 6 } $$

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

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

D
None of the above

Solution

Let $$A$$ be the event that Mr. $$A$$ will die within a year and $$B$$ be the event that Mr. $$B$$ will die within a year. Then, required probability
$$=P\left[ \left( A\quad will\quad die\quad and\quad B\quad alive \right) \quad or\quad \left( B\quad will\quad die\quad and\quad A\quad alive \right) \right] $$
$$=P\left[ \left( A\cap { B }^{ ' } \right) \cup \left( B\cap { A }^{ ' } \right) \right] $$
$$=P\left( A\cap { B }^{ ' } \right) +P\left( B\cap { A }^{ ' } \right)$$ [$$\because$$ these events are mutually exclusive]
$$=P\left( A \right) \cdot P\left( { B }^{ ' } \right) +P\left( B \right) \cdot P\left( { A }^{ ' } \right) $$ [$$\because A$$ and $$B$$ are independent]
$$=\dfrac { 1 }{ 2 } \left( 1-\dfrac { 1 }{ 3 } \right) +\dfrac { 1 }{ 3 } \left( 1-\dfrac { 1 }{ 2 } \right) $$
$$=\dfrac { 2 }{ 6 } +\dfrac { 1 }{ 6 } =\dfrac { 3 }{ 6 } =\dfrac { 1 }{ 2 } $$
Question 5
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The probabilities of solving a problem by three students $$A,B$$ and $$C$$ are $$\cfrac { 1 }{ 2 } ,\cfrac { 3 }{ 4 } $$ and $$\cfrac { 1 }{ 4 } $$ respectively. The probability that the problem will be solved is

A
$$\cfrac { 3 }{ 32 } $$

B
$$\cfrac { 3 }{ 16 } $$

C
$$\cfrac { 29 }{ 32 } $$

D
None of the above

Solution

Given, $$P(A)=\cfrac { 1 }{ 2 } ,P(B)=\cfrac { 3 }{ 4 } ,P(C)=\cfrac { 1 }{ 4 } $$
$$\Rightarrow P(\overline { A } )=\cfrac { 1 }{ 2 } ,P(\overline { B } )=\cfrac { 1 }{ 4 } ,P(\overline { C } )=\cfrac { 3 }{ 4 } $$
$$\therefore$$ Required probability $$=1-P(\overline { A } \cap \overline { B } \cap \overline { C } )=1-\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 4 } \times \cfrac { 3 }{ 4 } =\cfrac { 29 }{ 32 } $$
Question 6
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Out of $$50$$ tickets numbered $$00, 01, 02, ...., 49$$ one ticket is drawn randomly, the probability of the ticket having the product of its digits $$7$$ given that the sum of the digits is $$8$$, is

A
$$\dfrac {1}{14}$$

B
$$\dfrac {3}{14}$$

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

D
None of these

Solution

Total number of cases $$= ^{50}C_{1} = 50$$
Let $$A$$ be the event of selecting ticket with sum of digits $$'8'$$.
Favourable cases to $$A$$ are $$\left \{08, 17, 26, 35, 44\right \}$$.
Let $$B$$ be the event of selecting ticket with product of its digits $$'7'$$
Favourable cases to $$B$$ is only $$\left \{17\right \}$$.
Now, $$P(B/A) = \dfrac {P(A\cap B)}{P(A)}$$ $$= \dfrac {1/50}{5/50} = \dfrac {1}{5}$$
Question 7
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If $$A$$ and $$B$$ are two events such that $$P(A) = \dfrac {3}{4}$$ and $$P(B) = \dfrac {5}{8}$$, then

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

B
$$P(A'\cap B)\leq \dfrac{1}{4}$$

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

D
All of the above

Solution

$$A\subset A\cup B\Rightarrow P(A) \leq P(A\cup B)$$
$$\Rightarrow P(A\cup B) \geq 3/4$$
Also, $$P(A\cap B) = P(A) + P(B) - P(A\cup B) \geq P(A) + P(B) - 1$$
$$= \dfrac {3}{4} + \dfrac {5}{8} - 1 = \dfrac {3}{8}$$
Now, $$A\cap B\subset B\Rightarrow P(A \cap B)\leq P(B) = \dfrac {5}{8}$$
$$\therefore \dfrac {3}{8}\leq P(A\cap B) \leq \dfrac {5}{8} .... (i)$$
Next $$P(A\cap B') = P(A) - P(A\cap B)$$
$$\Rightarrow \dfrac {3}{4} - \dfrac {5}{8}\leq P(A\cap B') \leq \dfrac {3}{4} - \dfrac {3}{8}$$
$$\Rightarrow \dfrac {1}{8} \leq P(A\cap B') \leq \dfrac {3}{8}$$
$$\because P(A' \cap B) = P(B) - P(A\cap B)$$
$$\therefore P(A\cap B) = P(B) - P(A'\cap B')$$ [Using Eq. (i)]
$$\Rightarrow \dfrac {3}{8}\leq P(B) - P(A'\cap B)\leq \dfrac {5}{8}$$
$$\Rightarrow 0\leq P(A'\cap B)\leq \dfrac {1}{4}$$.
Question 8
1 / -0

If $$P(A) = 65, P(B) = 80$$, then $$P(A\cap B)$$ lies in the interval

A
$$[.30, .80]$$

B
$$[.35, .75]$$

C
$$[.4, .70]$$

D
$$[.45, .65]$$

Solution

$$P(A\cap B) \leq min \left \{P(A), P(B)\right \} = min\left \{.65, .80\right \} = 0.65$$

$$\therefore P(A\cap B) \leq .65$$

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

$$\geq .65 + .80 - 1 = 0.45$$

$$\therefore 0.45\leq P(A \cap B) \leq 0.65$$.
Question 9
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Let $$A$$ and $$B$$ be two events such that $$P (A \cup B) = P(A) + P(B)- P (A) P(B)$$. If $$0 < P(A) < 1$$ and $$0 < P (B) < 1$$, then $$P(A \cup B)'$$ is equal to

A
$$1 - P(A)$$

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

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

D
$$[1 - P(A)] P(B')$$

E
$$1$$

Solution

Given, $$P (A \cup B) = P(A) + P (B) - P(A) P(B)$$
Therefore, $$ P (A \cup B)' = 1 - P (A \cup B)$$
$$= 1 - [P (A) + P(B) - P(A) P(B)]$$
$$= 1 [1- P (A)] - P (B) [1 - P(A)]$$
$$= [1 - P (A)] [1 - P(B)]$$
Question 10
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If the probability of $$x$$ to fail in the examination is $$0.3$$ and that for $$Y$$ is $$0.2$$, then the probability that either $$X$$ or $$Y$$ fail in the examination is

A
$$0.5$$

B
$$0.44$$

C
$$0.6$$

D
None of these

Solution

Given, $$P(x)=0.3,P(y)=0.2$$
Now, $$P(x\cup y)=P(x)+P(y)-P(x\cap y)$$
Since, these are independent events, so
$$P(x\cap y)=P(x).P(y)$$
Thus, the required probability is $$0.3+0.2-0.06$$ $$=0.44$$.

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Re-Attempt Weekly Quiz Competition

Probability Test - 45

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Probability Test - 45

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$P\left( \overline { A } \cup \overline { B } \right) $$

$$P\left( A\cap \overline { B } \right) $$

$$P\left( \overline { A } \cap B \right) $$

$$P\left( A\cup B \right) $$

$$P\left( \overline { A } \cap \overline { B } \right) $$

Solution

Question 3

$$\dfrac {4}{21}$$

$$\dfrac {3}{21}$$

$$\dfrac {5}{21}$$

$$\dfrac {2}{21}$$

$$\dfrac {1}{21}$$

Solution

Question 4

$$\dfrac { 5 }{ 6 } $$

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

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

None of the above

Solution

Question 5

$$\cfrac { 3 }{ 32 } $$

$$\cfrac { 3 }{ 16 } $$

$$\cfrac { 29 }{ 32 } $$

None of the above

Solution

Question 6

$$\dfrac {1}{14}$$

$$\dfrac {3}{14}$$

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

None of these

Solution

Question 7

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

$$P(A'\cap B)\leq \dfrac{1}{4}$$

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

All of the above

Solution

Question 8

$$[.30, .80]$$

$$[.35, .75]$$

$$[.4, .70]$$

$$[.45, .65]$$

Solution

Question 9

$$1 - P(A)$$

$$1 - P(A')$$

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

$$[1 - P(A)] P(B')$$

$$1$$

Solution

Question 10

$$0.5$$

$$0.44$$

$$0.6$$

None of these

Solution

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