Probability Test

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Probability Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If A and B are two events such that $$ P\left ( A \right )=\dfrac{3}{8}, P\left ( B \right )= \dfrac{5}{8}$$ and $$P\left ( A\cup B \right )=\dfrac{3}{4}$$, then $$P\left ( A\cap \bar{B} \right )=$$

A
$$\dfrac{5}{8}$$

B
$$\dfrac{3}{8}$$

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

D
$$\dfrac{2}{8}$$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$\dfrac{3}{4}=\dfrac{3}{8}+\dfrac{5}{8}-P(A\cap B)$$
$$P(A\cap B)=\dfrac{1}{4}$$
$$P(A\cap \bar B)=P(A)-P(A\cap B)$$
$$P(A\cap \bar B)=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{8}$$
Question 2
1 / -0

If A and B are two mutually exclusive events such that $$P\left ( A \right )= 0.55$$ and $$P\left ( B \right ) = 0.35$$ then $$P\left ( \bar{A}\cup \bar{B}\right )= $$

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

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

C
1

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

Solution

Doesn't matter what is the value of $$P(A)$$ or $$P(B)$$.
Because A & B both are mutually cases.
So $$P\left ( \bar{A}\cup \bar{B}\right )=1$$.
Question 3
1 / -0

If A and B are two events such that $$P\left ( A \right )= \dfrac{1}{4}; P\left ( A\cup B \right )=\dfrac{1}{3}$$ and $$P\left ( B \right )= P$$ , the value of P if A and B are mutually exclusive is

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

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

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

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

Solution

A and B are mutually exclusive, so $$P(A\cap B)=0$$
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$\dfrac{1}{3}=\dfrac{1}{4}+P-0$$
$$P=\dfrac{1}{12}$$
Question 4
1 / -0

If A and B are two events such that $$P\left ( A \right )=\dfrac{1}{4}; P\left ( A\cup B \right )=\dfrac{1}{3}$$ and $$ P\left ( B \right )=p$$, the value of $$p$$ if A and B are independent is

A
$$\dfrac{1}{9}$$

B
$$\dfrac{2}{9}$$

C
$$\dfrac{4}{9}$$

D
$$\dfrac{5}{9}$$

Solution

If A and B are independent, then $$P\left (\dfrac{A}{B}\right )=P(A)$$
$$P\left ( \dfrac{A}{B}\right )$$$$=$$$$\dfrac{P(A \cap B)}{P(B)}$$

So, $$P(A \cap B)=P(A)P(B)=\dfrac{1}{4}p$$

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

$$\dfrac{1}{3}=\dfrac{1}{4}+p-\dfrac{1}{4}P$$

$$P=\dfrac{1}{9}$$
Question 5
1 / -0

If $$P\left ( A\cup B \right )= 0.8, P\left ( A\cap B \right )= 0.3$$, then $$P\left ( \bar{A} \right )+P\left ( \bar{B} \right )= $$

A
$$0.3$$

B
$$0.5$$

C
$$0.7$$

D
$$0.9$$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B)=1-P(\bar{A})+1-P(\bar{B})-P(A\cap B)$$
$$0.8=1-P(\bar{A})+1-P(\bar{B})-0.3$$
$$P(\bar{A})+P(\bar{B})=2-0.8-0.3=0.9$$
Question 6
1 / -0

The probability that at least one of the events $$A$$ and $$B$$ occurs is $$0.7$$ and they occur simultaneously with probability $$0.2.$$ Then $$P\left( \overline { A } \right) +P\left( \overline { B } \right) =$$

A
$$1.8$$

B
$$0.6$$

C
$$1.1$$

D
$$1.4$$

Solution

We have $$P\left( A\cup B \right) =0.7$$ and $$P\left( A\cap B \right) =0.2$$.
Now, $$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow P\left( A \right) +P\left( B \right) =0.9\Rightarrow 1-P\left( \overline { A } \right) +1-P\left( \overline { B } \right) =0.9\\ \Rightarrow P\left( \overline { A } \right) +P\left( \overline { B } \right) =1.1$$
Question 7
1 / -0

If A and B are two mutually exclusive events such that $$P\left ( A \right )= \frac{1}{2}P\left ( B \right )$$ and $$A\cup B= S$$ the sample space, then $$P\left ( A \right )$$

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

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

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

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

Solution

Given $$P\left( A \right) =\cfrac { 1 }{ 2 } P(B)$$ and $$A\cup B=S$$ , the sample space A and B are mutually exclusive events which means $$P\left( A\cap B \right) =0\\ $$
whereas $$P\left( A\cup B \right) =S$$ means $$P\left( A\cup B \right) =1$$
We know that $$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow 1=P\left( A \right) +2P\left( A \right) -0\\ \Rightarrow 1/3=P\left( A \right) $$
Question 8
1 / -0

A random variable $$X$$ has the probability distribution
$$X=x : \ \ \ \ \ \ \ \ 1 \quad,2\quad, 3\quad, 4\quad, 5\quad, 6\quad, 7\quad, 8$$
$$P(X) : \ \ \ 0.15, 0.23, 0.12, 0.1, 0.2, 0.08, 0.07, 0.05$$
Events $$E=$$ {X is a prime number} and
$$F= {X / X <4}$$
$$i:p\overline{(E\cup F )}=0.23$$
$$ii:p(\bar{E}\cup \bar{F})=0.65$$
Which of$$ I, II$$ is (are) true $$?$$

A
$$I$$ only

B
$$II$$ only

C
both $$I$$ and $$II$$

D
neither $$I$$ nor $$II$$

Solution

Prime numbers $$= 2,3,5,7$$
$$F = 1,2,3$$
$$E\cup F=\quad \{ 1,2,3,5,7\} \\ \overline { E } \cup \overline { F } =\{ 1,4,5,6,7,8\} \\ \overline { P(E\cup F) } =1-P(E\cup F)=0.1+0.08+0.05=0.23\\ P(\overline { E } \cup \overline { F } )=\quad 1-0.23-0.12=0.65$$
Hence, option 'C' is correct.
Question 9
1 / -0

Three numbers are chosen at random withoutreplacement from the set of integers $${1, 2, 3,....10}$$. The probability that the minimum of the chosen numbers is $$3$$ or the maximum of the chosen numbers is $$7$$, is equal to

A
$$23/120$$

B
$$13/120$$

C
$$13/60$$

D
$$11/40$$

Solution

Case 1: When 3 is minimum
Then the number of choices for choosing 2 cards other then $$7....4,5,6,8,9,10= 6C_2 = 15 $$
The number of ways when 7 is max but 3 not included: $$1,2,4,5,6=10$$ ways
When both 3,7 are present $$8$$ ways
Total desired cases $$33$$
Total ways of picking 3 numbers at random $$= 10C_3 = 120$$
Probability = $$33/120=11/40$$
Question 10
1 / -0

A and B are mutually exclusive events, then

A
$$P(A) \leq P(\overline{B})$$

B
$$P(\mathrm{A})>P(\overline{B})$$

C
$$\mathrm{P}(\mathrm{A})<\mathrm{P}(\mathrm{B})$$

D
$$\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})$$

Solution

If A and B are mutually exclusive this simply means$$\quad P \left( A\cap B \right) =\phi \\ \Rightarrow P\left( A\cup B \right) \leq 1\\ \Rightarrow P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \leq 1\quad \quad \quad \quad \quad \quad \quad P\left( A\cap B \right) =0\quad ,\because \left( A\cap B \right) =\phi \\ \Rightarrow P\left( A \right) \leq 1-P\left( B \right) \\ \Rightarrow P\left( A \right) \leq P\left( \overline { B } \right) $$

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Probability Test - 22

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Probability Test - 22

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SHARING IS CARING

Question 1

$$\dfrac{5}{8}$$

$$\dfrac{3}{8}$$

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

$$\dfrac{2}{8}$$

Solution

Question 2

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

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

1

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

Solution

Question 3

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

$$\dfrac{1}{6}$$

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

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

Solution

Question 4

$$\dfrac{1}{9}$$

$$\dfrac{2}{9}$$

$$\dfrac{4}{9}$$

$$\dfrac{5}{9}$$

Solution

Question 5

$$0.3$$

$$0.5$$

$$0.7$$

$$0.9$$

Solution

Question 6

$$1.8$$

$$0.6$$

$$1.1$$

$$1.4$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$$I$$ only

$$II$$ only

both $$I$$ and $$II$$

neither $$I$$ nor $$II$$

Solution

Question 9

$$23/120$$

$$13/120$$

$$13/60$$

$$11/40$$

Solution

Question 10

$$P(A) \leq P(\overline{B})$$

$$P(\mathrm{A})>P(\overline{B})$$

$$\mathrm{P}(\mathrm{A})<\mathrm{P}(\mathrm{B})$$

$$\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})$$

Solution

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