Probability Test

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Probability Test - 30

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Weekly Quiz Competition

Question 1
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A coin is tossed and a die is thrown simultaneously :
$$P$$ is the event of getting head and a odd number.
$$Q$$ is the event of getting either $$H$$ or $$T$$ and an even number.
$$R$$ is the event of getting a number on die greater than $$7$$ and a tail.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 12, n(P) = 3, n(Q) = 2, n(R) = 0$$

B
$$n(S) = 12, n(P) = 3, n(Q) = 3, n(R) = 0$$

C
$$n(S) = 12, n(P) = 3, n(Q) = 6, n(R) = 0$$

D
$$n(S) = 12, n(P) = 3, n(Q) = 5, n(R) = 0$$

Solution

A coin is tossed and a die is thrown simultaneously.
$$ S = \{ H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 \} $$
Thus $$ n (S) = 12$$
$$P$$ is the event of getting head and an odd number.
$$P = \{ H1, H3, H5 \}$$
Thus $$ n (P) = 3$$
$$Q$$ is the event of getting either $$H$$ or $$T$$ and an even number.
$$Q = \{ H2, H4, H6, T2, T4, T6 \}$$
Thus $$ n (Q) = 6$$
$$R$$ is the event of getting a number greater than $$7$$ and tail $$1$$.
$$R = \{ \}$$
$$\therefore n(R)=0$$
Hence, option C is correct.
Question 2
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Two dice are thrown :
$$P$$ is the event that the sum of the scores on the uppermost faces is a multiple of $$6$$.
$$Q$$ is the event that the sum of the scores on the uppermost faces is at least $$10$$.
$$R$$ is the event that same scores on both dice.
Which of the following pairs is mutually exclusive?

A
$$P,Q$$

B
$$P,R$$

C
$$Q,R$$

D
None of these

Solution

Possibilities of$$ P$$, $$(3,3), (6,6), (1,5), (5,1), (4,2),(2,4)$$
Possibilities of $$Q: (5,5), (5,6),(6,5), (6, 6)$$
Possibilities of $$R: (1,1), (2,2),(3, 3), (4,4), (5,5), (6,6)$$
Thus, the possibilities are neither exhaustive, nor mutually exclusive nor these are complementary probabilities.
Question 3
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Two die are thrown find the probability of getting the sum of the numbers on their upper faces divisible by 9.

A
$$\displaystyle \frac{1}{9}$$

B
$$\displaystyle \frac{2}{9}$$

C
$$\displaystyle \frac{1}{3}$$

D
$$\displaystyle \frac{4}{9}$$

Solution

Sample space of nos obtained is $$= 6\times6 $$ (As two dices are independent)
Sum of nos. obtained divisible by $$9 = {(3,6),(4,5)(5,4),(6,3)}=4$$
probability of getting the event = $$\cfrac{4}{36}$$= $$\cfrac{1}{9}$$
Question 4
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In a certain city two newspapers $$A$$ and $$B$$ are published. It is known that $$25$$% of the city population reads $$A$$ and $$20\%$$ of the population reads $$B$$. $$8\%$$ of the population reads both $$A$$ and $$B$$. It is known that $$30$$% of those who read $$A$$ but not $$ B $$ look into advertisements and $$40$$% of those who read $$B$$ but not $$A$$ look advertisements while $$50$$% of those who read both $$A$$ and $$B$$ look into advertisements . What is the percentage of the population who reads an advertisement?

A
$$\displaystyle \frac{139}{500}$$

B
$$\displaystyle \frac{361}{500}$$

C
$$\displaystyle \frac{139}{1000}$$

D
$$\displaystyle \frac{861}{1000}$$

Solution

Let $$P(A)$$ and $$P(B)$$ denote the precentage of city population who reda newspapers $$A$$ and $$B$$.
Then from given data ,we have
$$P(A)=25\%=$$$$\displaystyle =\frac{1}{4}$$,P(B)=20%=$$\displaystyle =\frac{1}{5}.$$
$$\displaystyle P\left ( A\cap B \right )=8%=\frac{2}{25}.$$
$$\displaystyle \therefore$$ Percentage of those who read A but not B $$\displaystyle P\left ( A\cap \bar{B} \right )=P\left ( A \right )-P\left ( A\cap B \right )$$ [see theorem 2 & 3] $$\displaystyle =\frac{1}{4}-\frac{2}{25}=\frac{17}{100}=17%$$
Similarly,$$\displaystyle P\left ( \bar{A}\cap B \right )=P\left ( B \right )-P\left ( A\cap B \right )$$ $$\displaystyle =\frac{1}{5}-\frac{2}{25}=\frac{3}{25}=12%$$
If $$P(C)$$ denotes the percentage of those who look into advertisement , then from the given data we obtain $$\displaystyle P\left ( C \right )30 %of P\left ( A\cap \bar{B} \right )+40%$$ of
$$\displaystyle P\left ( A\cap \bar{B} \right )+50
of P\left ( A\cap B \right )$$
$$\displaystyle =\frac{3}{10}\times \frac{17}{100}+\frac{2}{5}\times \frac{3}{25}+\frac{1}{2}\times \frac{2}{25}$$

$$\displaystyle =\frac{51+48+40}{1000}=\frac{139}{1000}=13.9\%$$.
Thus the percentage of population who read an advertisement is $$ 13.9$$%
Question 5
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Two fair die are thrown, find the probability that sum of the points on their uppermost faces is a perfect square or divisible by $$4$$:

A
$$\displaystyle \frac{12}{36}$$

B
$$\displaystyle \frac{13}{36}$$

C
$$\displaystyle \frac{15}{36}$$

D
$$\displaystyle \frac{16}{36}$$

Solution

two fair dice thrown
so total cases$$= 6\times6=36$$
and perfect square or divisible by $$4$$ that mean that sum should be equal to $$=4,8,9,12$$
for getting $$4= (1,3);(2,2);(3,1)$$
for getting $$8= (2,6);(3,5);(4,4);(5,3);(6,2)$$
for getting $$ 9= (3,6);(4,5);(5,4);(6,3)$$
for getting $$12=(6,6)$$
so total cases$$=3+5+4+1=13$$
so probability =$$\cfrac{13}{36}$$
Question 6
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The probability of an event A occurring is $$0.5$$ and of B occuring s $$0.3$$. If A and B are mutually exclusive events, then the probability of neither A nor B occurring is

A
$$0.6$$

B
$$0.5$$

C
$$0.7$$

D
none of these

Solution

$$\displaystyle P\left ( \bar{A}\cap \bar{B} \right )=P\left (

\overline{A\cup B} \right )$$ $$\displaystyle =1-P\left ( A\cup B \right

)=1-\left [ P\left ( A \right ) +P\left ( B \right )\right ]=1-0.5-0.3=0.2.$$
[ $$\displaystyle \because $$Events A, B are mutually exclusive ]
Question 7
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A coin is tossed twice. Events E and F are defined as follows :E=heads on first toss, F= heads on second toss.Find the probability of $$\displaystyle E\cup F.$$

A
Hence p $$\displaystyle(E\cup F) =\frac{1}{4}$$

B
Hence p $$\displaystyle(E\cup F) =\frac{3}{4}$$

C
Hence p $$\displaystyle(E\cup F) =\frac{1}{2}$$

D
none of these

Solution

Let H denotes head and T denotes tail .
Sample space $$S=\{(H, H) ,(H,T), (T, H), (T, T)\}$$
Total number of sample points $$n(S)=4$$
Now, $$ E = \{(H, H), (H, T)\} $$
and $$F=\{(H, H), (T, H)\}$$
$$E\cup F=\{(H, H), (H,T), (T, H)\}$$
So $$ n(E\cup F)=3$$
Hence $$\displaystyle P(E\cup F)=\displaystyle \frac{n(E\cup F)}{n(S)}=\frac{3}{4}$$
Question 8
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In a class of $$100$$ students, $$60$$ students drink tea, $$50$$ students drink coffee and $$30$$ students drink both. A student from class is selected at random, find the probability that student takes at least one of the two drinks (i.e. tea or coffee or both).

A
$$\displaystyle \frac{1}{5}$$

B
$$\displaystyle \frac{2}{5}$$

C
$$\displaystyle \frac{3}{5}$$

D
$$\displaystyle \frac{4}{5}$$

Solution

It is given that,
$$(1)$$ The total students in a class $$=100$$.
$$(2)$$ Students drinking coffee $$=50$$.
$$(3)$$ Students drinking tea $$=60$$.
$$(4)$$ Student drinking both coffee and tea $$=30$$.

Using (1), we get

$$n(S)=100$$

Let $$A$$ be the event that the selected student drink tea.
Using $$(3)$$, $$n(A)=60$$.

$$\therefore P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{100}=\dfrac{3}{5}$$.

Let $$B$$ be the event that the selected student drink coffee.

Using $$(2)$$, $$n(B)=50$$.

Therefore, $$ P(B)=\dfrac{n(B)}{n(S)}=\dfrac{50}{100}=\dfrac{1}{2}$$.

Now $$A \cap B$$ is the event that selected student drinks both coffee and tea and $$A\cup B$$ is the event that the student selected drinks either tea or coffee or both.

Now using $$(4)$$, $$n(A\cap B)=30$$.

$$\therefore P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{30}{100}=\dfrac{3}{10}$$.

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

$$=\dfrac {3}{5}+\dfrac {1}{2}-\dfrac {3}{10}$$

$$=\dfrac{6+5-3}{10}$$

$$=\dfrac{8}{10}=\dfrac{4}{5}$$

Thus $$P(A\cup B)=\dfrac{4}{5}$$

The probability that selected student takes at least one of the drinks is $$\dfrac{4}{5}$$.
Question 9
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If A and B are arbitrary events, then

A
$$\displaystyle P\left ( A\cap B \right )\geq P\left ( A \right )+P\left ( B \right )-1$$

B
$$\displaystyle P\left ( A\cap B \right )\leq P\left ( A \right )+P\left ( B \right )-1.$$

C
$$\displaystyle P\left ( A\cap B \right )= P\left ( A \right )+P\left ( B \right )-1.$$

D
none of these

Solution

For arbitrary events A, B, we have
$$\displaystyle P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right ).$$ ...(1)
Since probability of an event is less then or equal to 1, we have
$$\displaystyle P\left ( A\cup B \right )\leq 1$$ and so (1) gives
$$\displaystyle P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )\leq 1$$
or $$\displaystyle P\left ( A \right )+P\left ( B \right )-1\leq P\left ( A\cap B \right )$$
or $$\displaystyle P\left ( A\cap B \right )\geq P\left ( A \right )+P\left ( B \right )-1.$$
Question 10
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There are n different object 1, 2, 3, ... n distributed at random in n boxes $$\displaystyle A_{1},A_{2},A_{3},\cdots A_{N}.$$ Find the probability that two objects are placed in the boxes corresponding to their number.

A
$$1$$

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

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

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

Solution

Let $$\displaystyle E_{i}$$ be the event that ith object is placed in ithe box,
$$\displaystyle P\left ( E_{i} \right )=1.\dfrac{\left ( n-1 \right )}{n!}=\dfrac{1}{n},i=1,2\cdots n.$$
Now the probability that two boxes contain right objects.$$\displaystyle P\left ( E_{i}\cap E_{j} \right )=1.\dfrac{\left ( n-2 \right )!}{n!}$$
and these two objects can be out of n objects in $$\displaystyle ^{n}C_{2}$$ways.
So the probability that two objects are in right boxes.$$\displaystyle =^{n}C_{2}\dfrac{\left ( n-2 \right )!}{n!}=\dfrac{n!}{2!\left ( n-2 \right )!}.\dfrac{\left ( n-2 \right )}{n!}=\dfrac{1}{2}.$$

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Probability Test - 30

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Probability Test - 30

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Question 1

$$n(S) = 12, n(P) = 3, n(Q) = 2, n(R) = 0$$

$$n(S) = 12, n(P) = 3, n(Q) = 3, n(R) = 0$$

$$n(S) = 12, n(P) = 3, n(Q) = 6, n(R) = 0$$

$$n(S) = 12, n(P) = 3, n(Q) = 5, n(R) = 0$$

Solution

Question 2

$$P,Q$$

$$P,R$$

$$Q,R$$

None of these

Solution

Question 3

$$\displaystyle \frac{1}{9}$$

$$\displaystyle \frac{2}{9}$$

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

$$\displaystyle \frac{4}{9}$$

Solution

Question 4

$$\displaystyle \frac{139}{500}$$

$$\displaystyle \frac{361}{500}$$

$$\displaystyle \frac{139}{1000}$$

$$\displaystyle \frac{861}{1000}$$

Solution

Question 5

$$\displaystyle \frac{12}{36}$$

$$\displaystyle \frac{13}{36}$$

$$\displaystyle \frac{15}{36}$$

$$\displaystyle \frac{16}{36}$$

Solution

Question 6

$$0.6$$

$$0.5$$

$$0.7$$

none of these

Solution

Question 7

Hence p $$\displaystyle(E\cup F) =\frac{1}{4}$$

Hence p $$\displaystyle(E\cup F) =\frac{3}{4}$$

Hence p $$\displaystyle(E\cup F) =\frac{1}{2}$$

none of these

Solution

Question 8

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

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

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

$$\displaystyle \frac{4}{5}$$

Solution

Question 9

$$\displaystyle P\left ( A\cap B \right )\geq P\left ( A \right )+P\left ( B \right )-1$$

$$\displaystyle P\left ( A\cap B \right )\leq P\left ( A \right )+P\left ( B \right )-1.$$

$$\displaystyle P\left ( A\cap B \right )= P\left ( A \right )+P\left ( B \right )-1.$$

none of these

Solution

Question 10

$$1$$

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

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

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

Solution

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