Probability Test

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Probability Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

If A and B are arbitrary events, then

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

B
$$\displaystyle P\left ( A\cap B \right )\leq P\left ( A \right )+P\left ( B \right ),$$

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

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 a non-negative quantity, equation (1) gives
$$\displaystyle P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )\geq 0$$
or $$\displaystyle P\left ( A \right )+P\left ( B \right )\geq P\left ( A\cap B \right )$$
or $$\displaystyle P\left ( A\cap B \right )\leq P\left ( A \right )+P\left ( B \right ).$$
Question 2
1 / -0

An urn contains 11 balls numbered from 1 to 11. If a ball is selected at random, what is the probability of having a ball with a number which is mutliple of either 2 or 3 ?

A
$$\displaystyle \frac{7}{11}.$$

B
$$\displaystyle \frac{8}{11}.$$

C
$$\displaystyle \frac{4}{11}.$$

D
$$\displaystyle \frac{6}{11}.$$

Solution

Total number of balls $$=11$$
Let $$A$$ denote the event that the ball no. is a multiple of 2.
$$B$$ denote the event that the ball no. is a multiple of 3.
So, $$\displaystyle A=\left \{ 2, 4,6, 8, 10 \right \}$$
$$B=\left \{ 3, 6, 9 \right \}$$
$$\displaystyle \therefore A\cap B=\left \{ 6 \right \}$$
Thus $$\displaystyle P\left ( A\cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( A \cap B\right )$$
$$\displaystyle =\frac{5}{11}+\frac{3}{11}-\frac{1}{11}$$
$$\Rightarrow P(A\cup B)=\dfrac{7}{11}$$
Question 3
1 / -0

The probability that in the toss of two dice we obtain an even sum or a sum less than 5 is

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

B
$$\displaystyle \frac{1}{6}$$

C
$$\displaystyle \frac{2}{3}$$

D
$$\displaystyle \frac{5}{9}$$

Solution

Let $$A$$ be the event of obtaining an even sum and $$B$$ the event of obtaining a sum less than five, Then we have to find $$\displaystyle P\left ( A\cup B \right ).$$ Since $$A, B$$ are not mutually exclusive, we have
$$\displaystyle P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$$
$$\displaystyle = \frac{18}{36}+\frac{6}{36}-\frac{4}{36}= \frac{5}{9}.$$
since there are $$18$$ ways to get an even sum and $$6$$ ways to get a sum $$\displaystyle < 5,$$
viz. $$(1,3), (3,1), (2,2), (1,2), (2,1), (1,1) $$ and $$4$$ ways to get an even sum less than $$5$$, namely, $$(1,3), (3,1), (2,2), (1,1).$$
Question 4
1 / -0

A box contains $$3$$ red, $$3$$ white and $$3$$ green balls. A ball is selected at random. Find the probability that the ball picked up is a red ball:

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

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

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

D
$$\displaystyle \frac{3}{4}$$

Solution

Total number of outcomes $$= 9$$
Favorable outcomes (the ball is red) $$= 3$$
Probability $$=\dfrac{3}{9} = \dfrac{1}{3}$$
Question 5
1 / -0

A cricket team has 15 members, of whom only 5 can bowl. If the names of the 15 members are put into a hat and 11 drawn random, then the chance of obtaining an eleven containing at least 3 bowlers is

A
$$\displaystyle \frac{7}{13}$$

B
$$\displaystyle \frac{11}{15}$$

C
$$\displaystyle \frac{12}{13}$$

D
None of these.

Solution

Required probability $$\displaystyle =\frac{^{5}C_{3}\times ^{10}C_{8}}{^{15}C_{11}}+\frac{^{5}C_{4}\times ^{10}C_{7}}{^{15}C_{11}}+\frac{^{5}C_{5}\times ^{10}C_{6}}{^{15}C_{11}}$$
$$\displaystyle =\frac{1}{^{15}C_{11}}\left [ 10\times 45+5\times 120+1\times 210 \right ]$$
$$\displaystyle =\frac{1260\times 1\times 2\times 3\times 4}{15\times 14\times 13\times 12}=\frac{12}{13}.$$
Question 6
1 / -0

Two die are thrown. Find the probability of the event that the product of numbers on their upper faces is $$12$$:

A
$$\displaystyle \frac{1}{12}$$

B
$$\displaystyle \frac{1}{10}$$

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

D
$$\displaystyle \frac{1}{6}$$

Solution

Total number of outcomes $$= 36$$
Favourable outcomes $$(4,3), (3,4), (6,2), (2,6) = 4$$
Probability $$=\dfrac{4}{36} = \dfrac{1}{9}$$
Question 7
1 / -0

If $$A$$ and $$B$$ are mutually exclusive events, then $$\displaystyle P\left ( A\cap B \right )$$ equals

A
$$0$$

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

C
$$1$$

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

Solution

Given event $$A$$ and $$B$$ are mutually exclusive
$$\therefore A\cap B = \phi\Rightarrow n\cap B = 0\Rightarrow P\left(A\cap B\right) = 0$$
Question 8
1 / -0

The probability that at least one of the events A and B occurs is 0.6, If A and B occur simultaneously with probability 0.2, then $$\displaystyle P\left ( \bar{A} \right )+P\left ( \bar{B} \right )$$ is

A
0.4

B
0.8

C
1.2

D
1.4

Solution

We are given that $$\displaystyle P\left ( A\cup B \right )= 0.6,$$ and $$\displaystyle P\left ( A\cap B \right )= 0.2.$$
We know that if A and B are any two events, then
$$\displaystyle P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$$
or $$\displaystyle 0.6= 1-P\left ( \bar{A} \right )+1-P\left ( \bar{B} \right )-0.2$$
or $$\displaystyle P\left ( \bar{A} \right )+P\left ( \bar{B} \right )= 2-0.2-0.6= 1.2.$$
Question 9
1 / -0

A die is thrown. $$A$$ is the event that prime number comes up, $$B$$ is the event that the number divisible by three comes up, $$C$$ is the event that the perfect square number comes up. Then, $$A, B$$ and $$C$$ are :

A
Mutually exclusive

B
Mutually exhaustuve

C
Same

D
None of these

Solution

Here, the sample space is $$S=\{1,2,3,4,5,6\}$$
$$\Rightarrow n(S) = 6$$
$$A$$ : Getting a prime number $$\Rightarrow A=\{2,3,5\}$$
$$\therefore n(A) = 3$$
$$B$$ : Getting a number divisible by $$3 \Rightarrow B=\{3,6\}$$
$$\therefore n(B) = 2$$
$$C$$ : Getting a perfect square $$\Rightarrow C=\{1,4\}$$
$$\therefore n(C) = 2$$
Now, $$A\cup B\cup C=\{1,2,3,4,5,6\}$$
Thus, the events $$A,B$$ and $$C$$ are mutually exhaustive events.
Question 10
1 / -0

Two coins are tossed, $$A$$ is the event of getting at most one head, $$B$$ is the event getting both heads, $$C$$ is the event of getting same face on both the coins. The events $$A$$ and $$B$$ are:

A
Mutually exhaustive

B
Mutually exclusive

C
Same

D
None of these

Solution

$$n(S)\, =\, 4, n(A)\, =\, 3, n(B)\, =\, 1, n(C)\, =\, 2, A \cap B\, =\, \phi$$
$$\therefore\, A$$ and $$ B\,$$ are mutually exclusive events.

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Re-Attempt Weekly Quiz Competition

Probability Test - 31

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Probability Test - 31

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Question 1

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

$$\displaystyle P\left ( A\cap B \right )\leq P\left ( A \right )+P\left ( B \right ),$$

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

none of these

Solution

Question 2

$$\displaystyle \frac{7}{11}.$$

$$\displaystyle \frac{8}{11}.$$

$$\displaystyle \frac{4}{11}.$$

$$\displaystyle \frac{6}{11}.$$

Solution

Question 3

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

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{5}{9}$$

Solution

Question 4

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

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

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

$$\displaystyle \frac{3}{4}$$

Solution

Question 5

$$\displaystyle \frac{7}{13}$$

$$\displaystyle \frac{11}{15}$$

$$\displaystyle \frac{12}{13}$$

None of these.

Solution

Question 6

$$\displaystyle \frac{1}{12}$$

$$\displaystyle \frac{1}{10}$$

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

$$\displaystyle \frac{1}{6}$$

Solution

Question 7

$$0$$

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

$$1$$

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

Solution

Question 8

0.4

0.8

1.2

1.4

Solution

Question 9

Mutually exclusive

Mutually exhaustuve

Same

None of these

Solution

Question 10

Mutually exhaustive

Mutually exclusive

Same

None of these

Solution

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