Probability Test 16

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Probability Test 16

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Weekly Quiz Competition

Question 1
1 / -0

If $$A, B$$ and $$C$$ are mutually exclusive and exhaustive events, then $$P(A) + P(B) + P(C)$$ equals to -

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

B
$$1$$

C
$$0$$

D
Any value between $$0$$ and $$1$$

Solution

Given $$A,B,C$$ are mutually exclusive and exhaustive events
When $$A,B,C$$ are mutually exclusive events,
$$\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)$$
When $$A,B,C$$ are exheustive event,
$$\Rightarrow P(A)+P(B)+P(C)=1$$
Hence, the answer is probability always $$1.$$
Question 2
1 / -0

$$A, B$$ and $$C$$ are three mutually exclusive and exhaustive events such that $$P(A) = 2 P(B) = 3P(C)$$. What is $$P(B)$$?

A
$$\dfrac {6}{ 11}$$

B
$$\dfrac {6 }{ 22}$$

C
$$\dfrac {1 }{ 6}$$

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

Solution

Given $$A,B,C$$ are three mutually exclusive $$\xi$$ exhaustive events.
$$\Rightarrow P(A)=2P(B)=3P(C)$$
$$\Rightarrow P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) $$
$$=2P\left( B \right) +P\left( B \right) +\dfrac { 11P\left( B \right) }{ 3 } $$
$$=\dfrac { 6P\left( B \right) +3P\left( B \right) +2P\left( B \right) }{ 3 } =\dfrac { 11P\left( B \right) }{ 3 } $$
$$\Rightarrow P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) =1$$
$$\Rightarrow 1=\dfrac { 11P\left( B \right) }{ 3 }$$
$$\Rightarrow P\left( B \right)=\dfrac{3}{11}=\dfrac{6}{22}$$

Hence, the answer is $$\dfrac{6}{22}.$$
Question 3
1 / -0

An urn contains $$10$$ balls coloured either black or red When selecting two balls from the urn at random, the probability that a ball of each color is selected is $$8/15$$. Assuming that the urn contains more black balls then red balls, the probability that at least one black ball is selected, when selecting two balls, is

A
$$\dfrac {18}{45}$$

B
$$\dfrac {30}{45}$$

C
$$\dfrac {39}{45}$$

D
$$\dfrac {41}{45}$$

Solution

Number of black balls $$=x$$

Number of red balls $$=y$$

$$i)=x+y=10$$ (Total)

$$ii)\ P$$ (selecting exactly $$1$$ black & one red)

$$=^xC_1\times ^yC_1 /^{10}C_2=8/15$$

by equation : $$x^2-10x+24=0\ ;\ x=4$$ or $$6$$

since $$ x > y$$ it is $$6$$

$$iii)\ P$$ (selecting at least one black)

$$=^xC_1\times ^yC_1 +^xC_2 /^{10}C_2$$

Above equation reduced to $$\Rightarrow \ 2xy+x(x-1)/90$$

putting $$x=6$$, results is $$39/45$$
Question 4
1 / -0

Bag $$A$$ contains $$2$$ white and $$3$$ red balls and bag $$B$$ contains $$4$$ white and $$5$$ red balls. One ball is drawn at random from one of the bag is found to be red. Find the probability that it was drawn from bag $$B$$.

A
$$\dfrac{3}{8}$$

B
$$\dfrac{25}{52}$$

C
$$\dfrac{1}{8}$$

D
$$\dfrac{3}{14}$$

Solution

Let $$X$$ be the probability of choosing bag $$A$$,$$Y$$ be the probability of choosing bag $$B$$

Let $$E$$ be the probability of ball drawn is red

Then $$P\left( X \right) = \dfrac{1}{2}$$

$$P\left( Y \right) = \dfrac{1}{2}$$

$$P\left( {E/X} \right) = \dfrac{3}{5}$$

$$P\left( {E/Y} \right) = \dfrac{5}{9}$$

Apply the Bayes theorem:

$$P\left( {Y/E} \right) = \dfrac{{P\left( Y \right) \times P\left( {E/Y} \right)}}{{P\left( X \right) \times P\left( {E/X} \right) + P\left( Y \right) \times P\left( {E/Y} \right)}}$$

$$ = \dfrac{{\dfrac{1}{2} \times \dfrac{5}{9}}}{{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{5}{9}}}$$

$$ = \dfrac{{50}}{{104}}$$

$$ = \dfrac{{25}}{{52}}$$

Hence, the probability that the red ball is drawn from bag $$B$$ is $$ = \dfrac{{25}}{{52}}$$
Question 5
1 / -0

If $$P(A)=0.40,P(B)=0.35$$ and $$P\left( A\cup B \right) =0.55$$, then $$P(A/B)=$$ ____

A
$$\cfrac { 1 }{ 5 } $$

B
$$\cfrac { 8 }{ 11 } $$

C
$$\cfrac { 4 }{ 7 } $$

D
$$\cfrac { 3 }{ 4 } $$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B)=0.55$$
$$P(A)=0.40$$
$$P(B)=0.35$$
$$0.55=0.40+0.35-P(A\cap B)$$
$$P(A\cap B)=0.4+0.35-0.55=0.2$$
Now
$$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$$
$$P(A|B)=\dfrac{0.2}{0.35}$$
$$P(A|B)=\dfrac{4}{7}$$
Question 6
1 / -0

If $$A$$ and $$B$$ are mutually exclusive events, then

A
$$P(A) = P(A - B)$$

B
$$P(B) = P(A - B)$$

C
$$P(A) = P(A \cap B)$$

D
$$P(B) = P(A \cap B)$$

Solution

Given $$A$$ and $$B$$ are mutually exclusive
$$\Rightarrow P(A\cup B)=P(A)+(B)$$
$$\Rightarrow P(A-B)=P(A)-P(B)$$
When $$P(B)=0$$ i.e, $$P(A_B)+P(A)$$
$$\Rightarrow P(B)=0$$ is not a sure event.
Hence, the answer is $$P(A)=P(A-B).$$
Question 7
1 / -0

An experiment is known to be random if the results of the experiment -

A
Cannot be predicted

B
Can be predicted

C
Can be split into further experiments

D
Can be selected at random
Question 8
1 / -0

If A and B are two independent events such that $$P\left( A \right) = \dfrac{1}{2}$$ and $$P\left( B \right) = \dfrac{1}{5}$$, then

A
$$P\left( {A \cup B} \right) = \dfrac{3}{5}$$

B
$$P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{2}$$

C
$$P\left( {\dfrac{A}{{A\,\,\, \cup \,\,B}}} \right) = \dfrac{5}{6}$$

D
$$P\left( {\dfrac{{A\,\,\, \cap \,\,B}}{{\overline A \,\,\, \cup \,\,\overline B }}} \right) = 0$$

Solution

Given A and B are independent $$\Rightarrow P\left (A\cap B\right) =P(A) P(B) $$

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

$$\Rightarrow P(A\cup B) =P(A)+P(B)-P(A) P(B) $$

$$\Rightarrow P(A\cup B) =\dfrac {1}{2}+\dfrac {1}{5}-\dfrac {1}{10}$$

$$\Rightarrow P(A\cup B) =\dfrac {3}{5}$$
Question 9
1 / -0

Two unbiased dice are thrown. The probability that the sum of the numbers appearing on the top face of two dice is greater than $$7$$ if $$4$$ appear on the top face of the first dice is...

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

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

C
$$\dfrac {1}{12}$$

D
$$\dfrac {1}{6}$$

Solution

$$\begin{array}{l}\left. \begin{array}{l}{\rm{combinations}} = \left( {4,1} \right)\left( {4,2} \right)\\\left( {4,3} \right)\left( {4,4} \right)\end{array} \right\} \le 7\\\left. {\left( {4,5} \right)\left( {4,6} \right)} \right\} > 7\\{\rm{P}}\left( { > 7} \right) = \frac{1}{3}\end{array}$$
Question 10
1 / -0

There are six letters $$L_1, L_2, L_3, L_4, L_5, L_6$$ are their corresponding six envelopes $$E_1, E_2, E_3, E_4, E_5, E_6$$. Letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes, then number of arrangement equals?

A
$$6$$

B
$$9$$

C
$$44$$

D
$$4$$

Solution

There are three odd envelops and three even envelops
Favourable ways so that letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes are
For odd 1,3,5 -_,_,_
5 1 3,3 5 1
For even 2,4,6- _,_,_
6 2 4,4 6 2
There are four ways

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Re-Attempt Weekly Quiz Competition

Probability Test 16

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Probability Test 16

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SHARING IS CARING

Question 1

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

$$1$$

$$0$$

Any value between $$0$$ and $$1$$

Solution

Question 2

$$\dfrac {6}{ 11}$$

$$\dfrac {6 }{ 22}$$

$$\dfrac {1 }{ 6}$$

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

Solution

Question 3

$$\dfrac {18}{45}$$

$$\dfrac {30}{45}$$

$$\dfrac {39}{45}$$

$$\dfrac {41}{45}$$

Solution

Question 4

$$\dfrac{3}{8}$$

$$\dfrac{25}{52}$$

$$\dfrac{1}{8}$$

$$\dfrac{3}{14}$$

Solution

Question 5

$$\cfrac { 1 }{ 5 } $$

$$\cfrac { 8 }{ 11 } $$

$$\cfrac { 4 }{ 7 } $$

$$\cfrac { 3 }{ 4 } $$

Solution

Question 6

$$P(A) = P(A - B)$$

$$P(B) = P(A - B)$$

$$P(A) = P(A \cap B)$$

$$P(B) = P(A \cap B)$$

Solution

Question 7

Cannot be predicted

Can be predicted

Can be split into further experiments

Can be selected at random

Question 8

$$P\left( {A \cup B} \right) = \dfrac{3}{5}$$

$$P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{2}$$

$$P\left( {\dfrac{A}{{A\,\,\, \cup \,\,B}}} \right) = \dfrac{5}{6}$$

$$P\left( {\dfrac{{A\,\,\, \cap \,\,B}}{{\overline A \,\,\, \cup \,\,\overline B }}} \right) = 0$$

Solution

Question 9

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

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

$$\dfrac {1}{12}$$

$$\dfrac {1}{6}$$

Solution

Question 10

$$6$$

$$9$$

$$44$$

$$4$$

Solution

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