Probability Test

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Probability Test - 45

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Question 1
1 / -0

Directions For Questions

If it is given that A and B are two events such that P(B)$$= \frac{3}{5}$$ P(A/B) $$= \frac{1}{2}$$ and P(A $$\cup$$ B) $$= \frac{4}{5}$$,

...view full instructions

then P(A) equals

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

B
$$\frac{1}{5}$$

C
$$\frac{1}{2}$$

D
$$\frac{3}{5}$$

Solution

P(B)$$= \dfrac{3}{5}$$ P(A/B) $$= \dfrac{1}{2}$$ and P(A $$\cup$$ B) $$= \dfrac{4}{5}$$
$$\because P(A / B) = \dfrac{P(A \cap B)}{P(B)}$$
$$\Rightarrow \dfrac{1}{2} = \dfrac{P(A \cap B)}{3/5}$$
$$\Rightarrow P(A \cap B) = \dfrac{3}{5} \times \dfrac{1}{2} = \dfrac{3}{10}$$
And $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$\Rightarrow \dfrac{4}{5} = P(A) + \dfrac{3}{5} - \dfrac{3}{10}$$

$$\Rightarrow \dfrac{4}{5} = P(A) + \dfrac{3}{5} - \dfrac{3}{10}$$

$$\therefore P(A) = \dfrac{4}{5} - \dfrac{3}{5} + \dfrac{3}{10} = \dfrac{8 - 6 + 3}{10} = \dfrac{1}{2}$$

$$\Rightarrow P(A) = \dfrac{4}{5} - \dfrac{3}{5} + \dfrac{3}{10} = \dfrac{8 - 6 + 3}{10} = \dfrac{1}{2}$$
Question 2
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A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

A
$$\frac{45}{196}$$

B
$$\frac{135}{392}$$

C
$$\frac{15}{56}$$

D
$$\frac{15}{29}$$

Solution

Probability of getting exactly one red (R) ball
$$= P_R \cdot P_{\bar R} \cdot P_{\bar R} + P_{\bar R} \cdot P_{R} \cdot P_{\bar R} + P_{\bar R} \cdot P_{\bar R} \cdot P_R$$
$$= \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{2}{6} + \dfrac{3}{8} \cdot \dfrac{5}{7} \cdot \dfrac{2}{6} + \dfrac{3}{8} \cdot \dfrac{2}{7} \cdot \dfrac{5}{6}$$
$$= \dfrac{15}{4 \cdot 7 \cdot 6} + \dfrac{15}{4 \cdot 7 \cdot 6} + \dfrac{15}{4 \cdot 7 \cdot 6}$$
$$\dfrac{5}{56} + \dfrac{5}{56} + \dfrac{5}{56} = \dfrac{15}{56}$$
Question 3
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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 4
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A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement then the probability that two of the three balls were red, the first ball being red, is

A
$$\frac{1}{3}$$

B
$$\frac{4}{7}$$

C
$$\frac{15}{28}$$

D
$$\frac{5}{28}$$

Solution

Let $$E_1 =$$ Event that first ball being red
And $$E_2 =$$ Event that exactly two of three balls being red
$$\therefore P(E_1) = P_R \cdot P_{R} \cdot P_{R} + P_{R} \cdot P_{R} \cdot P_{\bar R} + P_{R} \cdot P_{\bar R} \cdot P_R + P_{R} \cdot P_{\bar R} \cdot P_{\bar R}$$

$$= \dfrac{5}{8} \cdot \dfrac{4}{7} \cdot \dfrac{3}{6} + \dfrac{5}{8} \cdot \dfrac{4}{7} \cdot \dfrac{3}{6} + \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{4}{6} + \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{2}{6}$$

$$\dfrac{60 + 60 + 60 + 30}{336} = \dfrac{210}{336}$$

$$P (E_1 \cap E_2) = P_R \cdot P_{\bar R} \cdot P_R + P_R \cdot P_R \cdot P_{\bar R}$$

$$= \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{4}{6} + \dfrac{5}{8} \cdot \dfrac{4}{7} \cdot \dfrac{3}{6} = \dfrac{120}{336}$$

$$\therefore P(E_2 / E_1) = \dfrac{P(E_1 \cap E_2)}{P(E_1)} = \dfrac{120/336}{210/336} = \dfrac{4}{7}$$
Question 5
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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 6
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A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

A
33/56

B
9/64

C
1/14

D
3/28

Solution
Question 7
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A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?

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

B
$$\frac{1} {2} \left(\frac{9} {10} \right)^{4}$$

C
$$\frac{1} {2} \left(\frac{9} {10} \right)^{5}$$

D
$$\left(\frac{9} {10} \right)^{5} + \frac{1} {2} \left(\frac{9} {10} \right)^{4}$$

Solution

Here, $$n = 5, p = \dfrac{10} {100} = \dfrac{1} {10} $$ and $$q = \dfrac{9} {10} $$
$$r \leq 1$$
$$\Rightarrow r = 0.1$$

Also, $$P(X =r) ={^n}C_rP^{r} q^{n=r}$$

$$\therefore P(X = r) = P(r = 0) + P(r = 1)$$

$$ = {^5}C_0 \left (\dfrac{1} {10} \right )^{0} \left (\dfrac{9} {10}  \right )^{5} +  {^5}C_1 \left (\dfrac{1} {10} \right )^{1} \left (\dfrac{9} {10}  \right )^{4}$$

$$ = \left (\dfrac{9} {10}  \right )^{5} +  5 \cdot \dfrac{1} {10} \cdot \left (\dfrac{9} {10}  \right )^{4}$$

$$ = \left (\dfrac{9} {10}  \right )^{5} + \dfrac{1} {2} \left (\dfrac{9} {10}  \right )^{4}$$
Question 8
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The probability distribution of a discrete random variable X is given below.

The value of K is

A
8

B
16

C
32

D
48

Solution

We know that, $$\sum { p }(x) =1$$
$$\Rightarrow \dfrac{5}{k}+\dfrac{7}{k}+\dfrac{9}{k}+\dfrac{11}{k}=1$$
$$\Rightarrow \dfrac{32}{k}=1$$
$$\therefore K = 32$$
Question 9
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The probability of guessing correctly at least 8 out 10 answers on a true-false type examination

A
7/64

B
7/128

C
45/1024

D
7/41

Solution

We know that, $$P(X=r)=^nC_r(p)^r(q)^{n-1}$$

Here, $$n=10,p=\dfrac{1}{2},q=\dfrac{1}{2}$$

And $$ r \geq 8 $$ i.e., $$r=8,9,10$$

$$\Rightarrow P(X=r)=P(r=8)+P(r=9)=P(r=10)$$

$$=^{10}C_8 \left(\dfrac{1}{2}\right)^8 \left(\dfrac{1}{2}\right)^{10-8} + ^{10}C_9 \left(\dfrac{1}{2}\right)^9 \left(\dfrac{1}{2}\right) + ^{10}C_{10} \left(\dfrac{1}{2}\right)^{10}$$

$$ \dfrac{10!}{8!2!} \left(\dfrac{1}{2}\right)^{10} + \dfrac{10!}{9!1!} \left(\dfrac{1}{2}\right)^{10}+\left(\dfrac{1}{2}\right)^{10}$$

$$=\left(\dfrac{1}{2}\right)^{10}. [45+10+1]=\left(\dfrac{1}{2}\right)^{10}\cdot 56$$

$$=\dfrac{1}{16\cdot16}\cdot56=\dfrac{7}{128}$$
Question 10
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In a box containing $$100$$ bulbs, $$10$$ are defective. The probability that out of a sample of $$5$$ bulbs, none is defective is

A
$$10^{-1}$$

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

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

D
$$\dfrac{9}{10}$$

Solution

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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

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

$$\frac{1}{5}$$

$$\frac{1}{2}$$

$$\frac{3}{5}$$

Solution

Question 2

$$\frac{45}{196}$$

$$\frac{135}{392}$$

$$\frac{15}{56}$$

$$\frac{15}{29}$$

Solution

Question 3

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

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

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

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

Solution

Question 4

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

$$\frac{4}{7}$$

$$\frac{15}{28}$$

$$\frac{5}{28}$$

Solution

Question 5

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 6

33/56

9/64

1/14

3/28

Solution

Question 7

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

$$\frac{1} {2} \left(\frac{9} {10} \right)^{4}$$

$$\frac{1} {2} \left(\frac{9} {10} \right)^{5}$$

$$\left(\frac{9} {10} \right)^{5} + \frac{1} {2} \left(\frac{9} {10} \right)^{4}$$

Solution

Question 8

8

16

32

48

Solution

Question 9

7/64

7/128

45/1024

7/41

Solution

Question 10

$$10^{-1}$$

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

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

$$\dfrac{9}{10}$$

Solution

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