Probability Test

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Probability Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

On tossing a fair coin for $$5$$ times, what is the probability that at least four of the five flips will be heads?

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

B
$$\cfrac{3}{10}$$

C
$$\cfrac{7}{10}$$

D
$$\cfrac{13}{16}$$

Solution

We have tossed coins for $$5$$ times.
Therefore the required probability using bernoulli's formula give us $$\dfrac{1}{2^{5}}(\:^{5}C_{4}+\:^{5}C_{5})=\dfrac{5+1}{32}=\dfrac{6}{32}=\dfrac{3}{16}$$
Question 2
1 / -0

Calculate the probability that a number selected at random from the set {$$2,3,7,12,15,22,72,108$$} will be divisible by both $$2$$ and $$3$$.

A
$$\cfrac{1}{4}$$

B
$$\cfrac{3}{8}$$

C
$$\cfrac{3}{5}$$

D
$$\cfrac{5}{8}$$

E
$$\cfrac{7}{8}$$

Solution

In the given set the number divisible by both $$2$$ and $$3$$ i.e numbers divisible by $$6$$ are $$\{12,72,108\}$$.
In total there are $$8$$ numbers in the sample set.
Therefore the required probability is $$\dfrac{3}{8}$$.
Question 3
1 / -0

For any two events $$A$$ and $$B$$, if $$\displaystyle P\left( A\cup B \right) =\dfrac {5}{6},\quad P\left( A\cap B \right) =\dfrac {1}{3},\quad P\left( B \right) =\dfrac {1}{2}$$, then $$\displaystyle P\left( A \right) $$ is:

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

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

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

D
none of these

Solution

Given, $$\displaystyle P\left( A\cup B \right)=\frac{5}{6} ,P\left( A\cap B \right) =\frac { 1 }{ 3 } ,P\left( B \right) =\frac { 1 }{ 2 } ,$$

$$\displaystyle \because \quad P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$

$$\displaystyle \therefore \quad P\left( A \right) =\frac { 5 }{ 6 } -\frac { 1 }{ 2 } +\frac { 1 }{ 3 } $$

$$\displaystyle =\frac { 5-3+2 }{ 6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } $$
Question 4
1 / -0

On a certain day the chance of rain is $$80\%$$ in Delhi and $$30\%$$ in Hyderabad. Assume that the chance of rain in the two cities is independent. What is the probability that it will not rain in either city?

A
$$7\%$$

B
$$14\%$$

C
$$24\%$$

D
$$50\%$$

Solution

Probability that it will not rain in Delhi $$= 100\% – 80\%$$
$$ = 20\% = 0.2$$
Probability that it will not rain in Hyderabad $$= 100\% – 30\% $$
$$= 70\% = 0.7$$
Since the chances of rain in both cities is independent
$$\therefore$$ Probability that it will not rain in either city $$=0.2 \times 0.7$$
$$= 0.14$$
$$\therefore$$ Probability that it will not rain in either city $$=14\%$$
Question 5
1 / -0

Four letters mailed today each have a probability of arriving in two days or sooner equal to $$\dfrac {2}{3}$$. Calculate the probability that exactly two of the four letters will arrive in two days or sooner.

A
$$\dfrac {4}{81}$$

B
$$\dfrac {16}{81}$$

C
$$\dfrac {6}{27}$$

D
$$\dfrac {8}{27}$$

E
$$\dfrac {4}{9}$$
Solution
- probability of arriving in two days or sooner is $$2/3$$
- probability of arriving after two days is $$1/3$$.
- the probability that only two of four letters arrive in two days or sooner is $${ ^{ 4 }{ C } }_{ 2 } \times 2/3 \times 2/3 \times 1/3 \times 1/3=6 \times 4/9 \times 1/9 = 8/27$$
Question 6
1 / -0

Two dice are tossed once. The probability of getting an even number at the first die or a total of $$8$$ is

A
$$\dfrac{1}{36}$$

B
$$\dfrac{3}{36}$$

C
$$\dfrac{11}{36}$$

D
$$\dfrac{20}{36}$$

Solution

$$A=\text{getting even no on Ist dice}$$
$$B=\text{getting sum 8}$$
So, $$n(A)=18$$
So, $$P(A\cup B)o=\dfrac{18+5-3}{36}=\dfrac{20}{36}$$
Question 7
1 / -0

If $$A, B$$ and $$C$$ are mutually exclusive and exclusive events of a random experiment such that $$P(B)=\dfrac{3}{2}P(A)$$ and $$P(C)=\dfrac{1}{2}P(B)$$, then $$P(A \cup C)=$$

A
$$\dfrac{10}{13}$$

B
$$\dfrac{3}{13}$$

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

D
$$\dfrac{7}{13}$$

Solution

We know $$P(A)+P(B)+P(C)=1$$

$$P(B)=\dfrac{3}{2}P(A)$$

$$P(C)=\dfrac{1}{2}\times P(B)=\dfrac{1}{2}\times \dfrac{3}{2}\times P(A)$$

$$P(C)=\dfrac{3}{4}P(A)$$

$$\Rightarrow P(A)+\dfrac{3}{2}P(A)+\dfrac{3}{4}P(A)=1$$

$$\Rightarrow P(A) = \dfrac{4}{13}$$

$$P(A\cup C)=P(A) + P(C)$$

$$=\dfrac{4}{13}+\dfrac{3}{4}\times \dfrac{4}{13}$$

$$P(A \cup C) = \dfrac{7}{13}$$
Question 8
1 / -0

Let $$S$$ be a set containing $$n$$ elements and we select two subsets $$A$$ and $$B$$ of $$S$$ at random, then the probability that $$A\cup B=S$$ and $$A\cap B=\phi$$, is.

A
$$2^n$$

B
$$n^2$$

C
$$\dfrac 1n$$

D
$$\dfrac {1}{2^n}$$

Solution

Given that $$S$$ contains $$n$$ elements and two sets $$A$$ and $$B$$ are selected.
Two sets $$A$$ and $$B$$ can be selected in $$2^{n}$$ ways.
The number of ways of selecting two sets such that their union is $$S$$ and intersection is nullset is $$1$$.
Therefore the probability is $$\frac{1}{2^{n}}$$.
So the correct option is $$D$$.
Question 9
1 / -0

Let $$A$$ and $$B$$ be events in a sample space $$S$$ such that $$P(A)=0.5, P(B)=0.4$$ and $$P(A\cup B)=0.6$$. Observe the following lists:
List I List II
i $$P(A\cap B)$$ a 0.4
ii $$P(A\cap \overline{B})$$ b 0.2
iii $$P(\overline{A}\cap B)$$ c 0.3
iv $$P(\overline{A}\cap \overline{B})$$ d 0.1
The correct match of List I from List II is

A
i- a, ii- b, iii- c, iv- d

B
i- c, ii- b, iii- d, iv- a

C
i- c, ii- b, iii- a, iv- d

D
i- c, ii- a, iii- b, iv- d

Solution

Given : $$P(A) = 0.5, P(B) = 0.4, P(A \cup B) = 0.6$$
i. $$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.6 = 0.3$$
ii. $$P(A \cap \overline{B} ) = P(A) - P(A \cap B) = 0.5 - 0.3 = 0.2$$
iii. $$P(\overline{A} \cap B) = P(B) - P(A \cap B) = 0.4 - 0.3 = 0.1$$
iv $$P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$$
Question 10
1 / -0

In a sample survey of $$640$$ people, it was found that $$400$$ people have a secondary school certificate. If a person is selected at random, what is the probability that the person does not have such certificate?

A
$$0.375$$

B
$$0.625$$

C
$$0.725$$

D
$$0.875$$

Solution

In a sample survey of 640 people, it was found that 400 people have a secondary school certificate.

person does not have such certificate$$640-400=240$$
Therefore the probability that the person does not have such certificate $$P(E)=\dfrac{240}{640}=0.375$$

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Re-Attempt Weekly Quiz Competition

	List I		List II
i	$$P(A\cap B)$$	a	0.4
ii	$$P(A\cap \overline{B})$$	b	0.2
iii	$$P(\overline{A}\cap B)$$	c	0.3
iv	$$P(\overline{A}\cap \overline{B})$$	d	0.1

Probability Test - 38

Result

Probability Test - 38

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SHARING IS CARING

Question 1

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

$$\cfrac{3}{10}$$

$$\cfrac{7}{10}$$

$$\cfrac{13}{16}$$

Solution

Question 2

$$\cfrac{1}{4}$$

$$\cfrac{3}{8}$$

$$\cfrac{3}{5}$$

$$\cfrac{5}{8}$$

$$\cfrac{7}{8}$$

Solution

Question 3

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

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

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

none of these

Solution

Question 4

$$7\%$$

$$14\%$$

$$24\%$$

$$50\%$$

Solution

Question 5

$$\dfrac {4}{81}$$

$$\dfrac {16}{81}$$

$$\dfrac {6}{27}$$

$$\dfrac {8}{27}$$

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

Solution

Question 6

$$\dfrac{1}{36}$$

$$\dfrac{3}{36}$$

$$\dfrac{11}{36}$$

$$\dfrac{20}{36}$$

Solution

Question 7

$$\dfrac{10}{13}$$

$$\dfrac{3}{13}$$

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

$$\dfrac{7}{13}$$

Solution

Question 8

$$2^n$$

$$n^2$$

$$\dfrac 1n$$

$$\dfrac {1}{2^n}$$

Solution

Question 9

i- a, ii- b, iii- c, iv- d

i- c, ii- b, iii- d, iv- a

i- c, ii- b, iii- a, iv- d

i- c, ii- a, iii- b, iv- d

Solution

Question 10

$$0.375$$

$$0.625$$

$$0.725$$

$$0.875$$

Solution

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