Probability Test

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Probability Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

A number is chosen at random from among the 1st 50 natural numbers. The probability that the number chosen is either a prime number or a multiple of 5 is

A
$$\displaystyle \frac{12}{25}$$

B
$$\displaystyle \frac{1}{2}$$

C
$$\displaystyle \frac{14}{25}$$

D
$$\displaystyle \frac{1}{4}$$

Solution

P(either prime no. or multiple of 5) $$=$$ P(prime no.)$$+$$P(multiple 5)$$ -$$ P(both prime & multiple of 5)
$$= (15+10-1)/50 = 12/25$$
Question 2
1 / -0

A lot contains $$50$$ defective and $$50$$ non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events $$A, B, C$$ are defined as $$A = $$ {the first bulb is defective}, $$B =$$ {the second bulb is non defective}, $$C =$$ {the two bulbs are both defective or both non defective}, then which of the following statements is/are true?(1) $$A, B, C$$ are pair wise independent.(2) $$A, B, C$$ are independent.

A
Only (1) is true.

B
Both (1) and (2) are true.

C
Only (2) is true.

D
Both (1) and (2) are false.

Solution

Let's denote the defective item by $$D$$ and non-defective by $$\overline { D } $$

$$P\left( A \right) =P\left( D \right) \left\{ P\left( \overline { D } \right) +P\left( D \right) \right\} =\dfrac { 50 }{ 100 } \times 1=\dfrac { 1 }{ 2 }$$

$$P\left( B \right) =\left\{ P\left( \overline { D } \right) +P\left( D \right) \right\} \times P\left( \overline { D } \right) =1\times \dfrac { 50 }{ 100 } =\dfrac { 1 }{ 2 }$$

$$ P\left( C \right) =P\left( DD\cup \overline { D } \overline { D } \right) =P\left( DD \right) +P\left( \overline { D } \overline { D } \right) -P\left( DD\cap \overline { D } \overline { D } \right)$$

$$\Rightarrow P\left( C \right) =\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } -0=\dfrac { 1 }{ 2 }$$

Now
$$P\left( A\cap B \right) =P\left( D\overline { D } \right) =\dfrac { 50 }{ 100 } \times \dfrac { 50 }{ 100 } =\dfrac { 1 }{ 4 }$$

$$P\left( B\cap C \right) =P\left( \overline { D } \overline { D } \right) =\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } =\dfrac { 1 }{ 4 }$$

$$P\left( C\cap A \right) =P\left( DD \right) =\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } =\dfrac { 1 }{ 4 } $$

Thus, we can see that
$$P\left( A\cap B \right) =P\left( A \right) P\left( B \right) ,P\left( B\cap C \right) =P\left( B \right) P\left( C \right)$$

$$P\left( C\cap A \right) =P\left( C \right) P\left( A \right) $$
Question 3
1 / -0

Given two events A and B, if the odds against Aare 2 to 1, and those in favour of A$$\cup $$B are 3 to1, then

A
$$\displaystyle \frac{1}{3}\leq P(B)\leq\frac{1}{2}$$

B
$$\displaystyle \frac{1}{2}\leq P(B)\leq \frac{3}{4}$$

C
$$\displaystyle \frac{5}{12}\leq P(B)\leq\frac{3}{4}$$

D
$$0\leq P(B)\leq 1$$

Solution

$$Odds\quad against\quad A=P(A^ c)/P(A)=\dfrac { 2 }{ 1 } \\ \Rightarrow \{ 1-P(A)\} /P(A)=2\quad \\ \Rightarrow 1/P(A)-1=2\\ \Rightarrow 1/P(A)=3\\ \Rightarrow P(A)=1/3 \\ \\ Also\quad Odds\quad in\quad favour\quad of\quad A\cup B=\dfrac { P\left( A\cup B \right) }{ P\left( A\cup B \right) ^ c } =\dfrac { 3 }{ 1 } \\ \Rightarrow \dfrac { P\left( A\cup B \right) }{ 1-P\left( A\cup B \right) } =3\\ \Rightarrow \dfrac { 1-P\left( A\cup B \right) }{ P\left( A\cup B \right) } =\dfrac { 1 }{ 3 } \\ \Rightarrow \dfrac { 1 }{ P\left( A\cup B \right) } =1+\dfrac { 1 }{ 3 } =\dfrac { 4 }{ 3 } \\ \Rightarrow P\left( A\cup B \right) =\dfrac { 3 }{ 4 }$$

since we know that $$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right)$$

Hence putting the values of results obtained above we get $$\Rightarrow \dfrac { 3 }{ 4 } =\dfrac { 1 }{ 3 } +P\left( B \right) -P\left( A\cap B \right)$$
on rearranging the terms we get $$\Rightarrow P\left( B \right) =P\left( A\cap B \right) +\dfrac { 5 }{ 12 } $$
$$P\left( A\cap B \right) \quad can\quad never\quad be\quad equal\quad to\quad P\left( B \right) \\ \Rightarrow \quad \dfrac { 5 }{ 12 } \le P\left( B \right) \le \dfrac { 5 }{ 12 } +P\left( A \right) $$

$$ 0\le P\left( A\cap B \right) \le P\left( A \right) \\ \Rightarrow \dfrac { 5 }{ 12 } \le P\left( B \right) \le \dfrac { 3 }{ 4 } $$
Question 4
1 / -0

Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:

No. of defective bulbs
0
1
2
3
4
5
6
more than 6
frequency
400
180
48
41
18
8
3
2

One carton was selected at random. The probability of defective bulbs being less than 4 is

A
$$\dfrac{89}{700}$$

B
$$\dfrac{432}{700}$$

C
$$\dfrac{600}{700}$$

D
$$\dfrac{669}{700}$$

Solution

$$P(\text {carton having defective bulbs less than 4})$$

$$ = 1-P(\text {carton having defective bulbs greater than or equal to 4})$$

$$= 1-\cfrac {18+8+3+2}{700} = 1-\cfrac {31}{700} = \cfrac {669}{700}$$
Question 5
1 / -0

The probability of raining on day $$1$$ is $$0.2$$ and on day $$2$$ is $$0.3$$. The probability of raining on both the days is

A
$$0.2$$

B
$$0.1$$

C
$$0.06$$

D
$$0.25$$

Solution

$${\textbf{Step -1: Calculate the probability of raining on both day.}}$$
$${\text{The probability of raining on day1, P(1) = 0}}{\text{.2}}$$
$${\text{The Probability of raining on day 2, P(2) = 0}}{\text{.3}}$$

$${\text{Raining on both day are independent of each other.}}$$

  $${\text{So, the probability of raining on both days = P}}\left( {1 \cap 2} \right)$$
$$ = {\text{P(1)}}{\text{.P(2)}}$$

  $$ = 0.2 \times 0.3$$

  $$ = 0.06$$

$$\textbf{Hence, option C is correct.}$$
Question 6
1 / -0

The probability that atleast one of the events $$A, B$$ happens is $$0.6$$. If probability of their simultaneously happening is $$0.5$$, then $$P(\bar{A})+P(\bar{B})=$$

A
$$0.4$$

B
$$0.8$$

C
$$0.9$$

D
$$1.4$$

Solution

$$P(A\cup B)= 0.6\\ P(A\cap B)= 0.5$$
$$ P(\overline { A } )+P(\overline { B } )=1-P(A)+1-P(B) $$
$$ =2-\{ P(A\cup B)+P(A\cap B)\}= 0.9$$
Question 7
1 / -0

Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:

No. of defective bulbs
0
1
2
3
4
5
6
more than 6
frequency
400
180
48
41
18
8
3
2

One carton was selected at random. What is the probability that it has more than 1 defective bulbs?

A
$$\dfrac{60}{350}$$

B
$$\dfrac{67}{350}$$

C
$$\dfrac{71}{350}$$

D
$$\dfrac{77}{350}$$

Solution

To find the number of cartons with no. of defective bulbs from 2 to 6, we can remove the number of cartons with no. of defective bulbs 0 or 1 from the whole set

No. of cartoons with 0 or 1 defective bulb $$= 400+180 = 580$$

So, $$P\text {(carton with 2 to 6 defective bulbs)}$$

$$ = 1-P\text{(cartoon with 0 or 1 defective bulb)}$$

$$=1-\cfrac {580}{700} $$

$$= \cfrac {120}{700} $$

$$= \cfrac {60}{350}$$
Question 8
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The probabilites of three events $$A,B$$ and $$C$$ are $$P\left( A \right) =0.6,P\left( B \right) =0.4$$ and $$P\left( C \right) =0.5$$. If $$P\left( A\cup B \right) =0.8,P\left( A\cap C \right) =0.3,P\left( A\cap B\cap C \right) =0.2$$ and $$P\left( A\cup B\cup C \right) \ge 0.85$$, then

A
$$0.2\le P\left( B\cap C \right) \le 0.35$$

B
$$0.5\le P\left( B\cap C \right) \le 0.85$$

C
$$0.1\le P\left( B\cap C \right) \le 0.35$$

D
None of these

Solution

Given $$P\left( A \right) =0.6,P\left( B \right) =0.4$$ and $$P\left( C \right) =0.5$$. If $$P\left( A\cup B \right) =0.8,P\left( A\cap C \right) =0.3,P\left( A\cap B\cap C \right) =0.2$$ and $$P\left( A\cup B\cup C \right) \ge 0.85$$

$$P\left( A\cap B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cup B\right) =0.6+0.4-0.8=0.2$$

$$P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap C \right) +P\left( A\cap B\cap C \right) -P\left( A\cap B \right) -P\left( B\cap C \right) $$

$$P\left( A\cup B\cup C \right) =0.6+0.4 +0.5 -0.3+0.2 -0.2 -P\left( B\cap C \right) $$

$$\Rightarrow P\left( B\cap C \right) =1.2-P\left( A\cup B\cup C \right) $$ ...(1)

$$\because 0.85\le P\left( A\cup B\cup C \right) \le 1$$

$$ \Rightarrow P\left( B\cap C \right) \le 1.2-0.85$$ [From (1)]

and $$P\left( B\cap C \right) \ge 1.2-1\Rightarrow 0.2\le P\left( B\cap C \right) \le 0.35.$$
Question 9
1 / -0

If events $$A$$ and $$B$$ are independent and $$P(A)=0.15, P(A\cup B)=0.45$$, then $$P(B)=$$

A
$$\dfrac {6}{13}$$

B
$$\dfrac {6}{17}$$

C
$$\dfrac {6}{19}$$

D
$$\dfrac {6}{23}$$

Solution

Given, $$P(A) = 0.15$$, $$P(A \cup B)$$ = 0.45
We have $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
and $$P(A \cap B) = P(A).P(B)$$
Therefore, $$0.45 = 0.15 +P(B) - 0.15 P(B) $$
$$\Rightarrow 0.30 = 0.85 P(B)$$
$$\Rightarrow P(B) =$$ $$\dfrac{30}{85}$$ $$=$$ $$\dfrac{6}{17}$$
Question 10
1 / -0

If $$\displaystyle \frac{1-3p}{2},\frac{(1+4p)}{3},\frac{1+p}{6}$$ are the probabilities of three mutually exclusive and exhaustive events,then the set of all values of p is

A
$$(0,1)$$

B
$$[-\displaystyle \frac{1}{4},\frac{1}{3}]$$

C
$$[0,\displaystyle \frac{1}{3}]$$

D
$$(0,\infty)$$

Solution

For mutually exclusive and exhaustive events,
$$ \displaystyle 0\leq\frac{1-3p}{2}\leq 1,0\leq \frac{1+4p}{3}\leq 1 ,0\leq \frac{1+p}{6}\leq 1$$ and $$\displaystyle 0\leq \frac{1-3p}{2}+ \frac{1+4p}{3}+\frac{1+p}{6} \leq 1$$
$$\Rightarrow \displaystyle 0\leq 1-3p \leq 2,0\leq 1+4p\leq 3 ,0\leq 1+p\leq 6$$ and $$\displaystyle 0\leq \frac{3(1-3p)+2(1+4p)+1+p}{6} \leq 1$$
$$\Rightarrow \displaystyle -1\leq -3p \leq 1,-1\leq 4p\leq 2 ,-1\leq p\leq 5$$ and $$\displaystyle 0\leq 1\leq

1$$
$$\Rightarrow \displaystyle -\frac{1}{3}\leq p \leq \frac{1}{3} ,-\frac{1}{4}\leq p\leq \frac{1}{2} ,-1\leq p\leq 5$$
Now taking intersection of above we get $$p\in \left[-\cfrac{1}{4}, \cfrac{1}{3}\right]$$

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No. of defective bulbs	0	1	2	3	4	5	6	more than 6
frequency	400	180	48	41	18	8	3	2

Probability Test - 24

Result

Probability Test - 24

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Question 1

$$\displaystyle \frac{12}{25}$$

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

$$\displaystyle \frac{14}{25}$$

$$\displaystyle \frac{1}{4}$$

Solution

Question 2

Only (1) is true.

Both (1) and (2) are true.

Only (2) is true.

Both (1) and (2) are false.

Solution

Question 3

$$\displaystyle \frac{1}{3}\leq P(B)\leq\frac{1}{2}$$

$$\displaystyle \frac{1}{2}\leq P(B)\leq \frac{3}{4}$$

$$\displaystyle \frac{5}{12}\leq P(B)\leq\frac{3}{4}$$

$$0\leq P(B)\leq 1$$

Solution

Question 4

$$\dfrac{89}{700}$$

$$\dfrac{432}{700}$$

$$\dfrac{600}{700}$$

$$\dfrac{669}{700}$$

Solution

Question 5

$$0.2$$

$$0.1$$

$$0.06$$

$$0.25$$

Solution

Question 6

$$0.4$$

$$0.8$$

$$0.9$$

$$1.4$$

Solution

Question 7

$$\dfrac{60}{350}$$

$$\dfrac{67}{350}$$

$$\dfrac{71}{350}$$

$$\dfrac{77}{350}$$

Solution

Question 8

$$0.2\le P\left( B\cap C \right) \le 0.35$$

$$0.5\le P\left( B\cap C \right) \le 0.85$$

$$0.1\le P\left( B\cap C \right) \le 0.35$$

None of these

Solution

Question 9

$$\dfrac {6}{13}$$

$$\dfrac {6}{17}$$

$$\dfrac {6}{19}$$

$$\dfrac {6}{23}$$

Solution

Question 10

$$(0,1)$$

$$[-\displaystyle \frac{1}{4},\frac{1}{3}]$$

$$[0,\displaystyle \frac{1}{3}]$$

$$(0,\infty)$$

Solution

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