Probability Test

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Probability Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

In a box containing $$100$$ are defective. The probability that out of a sample of $$5$$ bulbs, none is defective is:

A
$${ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }$$

B
$${10}^{-1}$$

C
$$\cfrac{9}{10}$$

D
$${ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$$

Solution

Number of bulbs in box $$=100$$
Number of defective bulbs $$=10$$
Probability of defective bulbs $$=\cfrac { 10 }{ 100 } =\cfrac { 1 }{ 10 } $$
So probability of non defective bulb$$1-\cfrac { 1 }{ 10 } =\cfrac { 9 }{ 10 } $$
Probability that no bulb is defective out a sample of $$5$$
bulbs $$P(X=0)={ _{ }^{ 5 }{ C } }_{ 0 }{ \left( \cfrac { 9 }{ 10 } \right) }^{ 5-0 }{ \left( \cfrac { 1 }{ 10 } \right) }^{ 0 }={ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$$
Question 2
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One ticket is selected randomly from the set of $$100$$ tickets numbered as $$\left\{00, 01, 02, 03,04,05,...,98, 99\right\}$$. $$E_1$$ and $$E_2$$ denote the sum and product of the digits of the number of the selected ticket. The value of $$P \left(\displaystyle \frac{E_{1}=9}{E_{2}=0} \right)$$ is

A
$$\dfrac{1}{19}$$

B
$$\dfrac{2}{19}$$

C
$$\dfrac{3}{19}$$

D
$$\dfrac{1}{18}$$

Solution

A ticket can be selected in $$100$$ ways
$$E_2=$$ product of digits is zero, which is possible when number is $$\left\{00,01,02,...,10.20.30.40.60.70.80.90\right\}$$
$$\displaystyle N\left( { E }_{ 2 } \right) =19\Rightarrow P\left( { E }_{ 2 } \right) =\frac { 19 }{ 100 } $$
$$E_1=$$ sum of digit is $$9\Rightarrow { E }_{ 1 }\cap { E }_{ 2 }=\left\{ 09,90 \right\} $$
So $$n\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =2$$
$$\displaystyle \therefore P\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =\frac { 2 }{ 100 } $$
Hence $$\displaystyle P\left( \frac { { E }_{ 1 }=9 }{ { E }_{ 2 }=0 } \right) =\frac { 2 }{ 19 } $$
Question 3
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A box contains $$100$$ tickets numbered $$1, 2, ...,100$$. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than $$10$$. The probability that the minimum number on them is $$5$$ is

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

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

C
$$\displaystyle \frac{3}{9}$$

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

Solution

We need to find the probability of a number being $$5$$ given that the maximum number is not more than $$10$$.Thus, we need to find $$P(\dfrac { minimum=5 }{ maximum\le 10 } )=\dfrac { P((minimum=5)\cap (maximum\le 10)) }{ P(maximum\le 10) }$$

Hence, probability = $$\dfrac { \dfrac { ^{ 5 }{ C_1 } }{ ^{ 100 }{ C_2 } } }{ \dfrac { ^{ 10 }{ C_2 } }{ ^{ 100 }{ C_2 } } } =\quad \dfrac { 5 }{ 45 } =\dfrac { 1 }{ 9 } $$
Question 4
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lt is given that the events $$\mathrm{A}$$ and $$\mathrm{B}$$ are such that $$P(A)=\displaystyle \frac{1}{4}, P(\displaystyle \frac{A}{B})=\frac{1}{2}$$ and $$P(\displaystyle \frac{B}{A})=\frac{2}{3}$$ then $$\mathrm{P}(\mathrm{B})=$$

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

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

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

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

Solution

$$P(\dfrac{B}{A})=\dfrac{P(A \cap B)}{P(A)} $$
Using this formula, we get $$P(A \cap B)= \dfrac{1}{6}$$
Now $$P(\dfrac{A}{B})=\dfrac{P(A \cap B)}{P(B)} $$

then, $$ P(B)= \dfrac{1}{3}$$
Question 5
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A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that it contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one, without replacement and tested till all defective articles are found. The probability that the testing procedure ends at the 12th testing is

A
$$\displaystyle \frac{11}{1900}$$

B
$$\displaystyle \frac{44}{1900}$$

C
$$\displaystyle \frac{66}{1900}$$

D
$$\displaystyle \frac{99}{1900}$$

Solution

$$\textbf{Step -1: Find the probability of finding all the defectives in the twelfth testing in both cases.}$$

$$\text{Let }A_1=\text{ the event that the lot contain 2 defective articles.}$$

  $$A_2=\text{ the event that the lot contains 3 defective articles.}$$

  $$A=\text{ the event that the testing procedure ends with the twelfth testing.}$$

  $$\text{It is given that,}$$

  $$P(A_1)=0.4\text{ and }P(A_2)=0.6$$

  $$\text{The testing procedure ending at the twelfth testing means that all the defective articles must be}$$

    $$\text{found in the first eleven testing except one which will be found at the twelfth testing.}$$

  $$\text{In case the lot contains 2 defective articles,}$$

  $$P(A|A_1)=\dfrac{^{2}C_{2}\times ^{18}C_{10}}{^{20}C_{11}}\times\dfrac{1}{9}$$

  $$\text{Similarly, in case the lot contains 3 defective articles,}$$

  $$P(A|A_2)=\dfrac{^{3}C_2\times^{17}C_7}{^{20}C_{11}}\times\dfrac{1}{9}$$

$$\textbf{Step -2: Find the required probability.}$$

   $$\text{The required probability,}$$

  $$=P(A)=P(A∩A_1)+P(A∩A_2)$$

  $$=P(A_1)\times P(A|A_1)+P(A_2)\times P(A|A_2)$$

  $$=0.4\times\dfrac{^{2}C_{2}\times ^{18}C_{10}}{^{20}C_{11}}\times\dfrac{1}{9}+0.6\times\dfrac{^{3}C_2\times^{17}C_7}{^{20}C_{11}}\times\dfrac{1}{9}$$

  $$=0.4\times\dfrac{11}{190}+0.6\times\dfrac{11}{228}$$

  $$=\dfrac{44}{1900}+\dfrac{66}{2280}$$

  $$=\dfrac{99}{1900}$$

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

A letter is known to have come form TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. The probability that the letter has come from CALCUTTA is

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

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

C
$$\displaystyle \frac{1}{22}$$

D
$$\displaystyle \frac{21}{22}$$

Solution

Let's calculate the total probability of getting $$TA$$
$$P\left( TA \right) =P\left( TA/TATANAGAR \right) \times P\left( TATANAGAR \right) +P\left( TA/CALCUTTA \right) \times P\left( CALCUTTA \right) \\ TATANAGAR\quad are\quad TA,AT,TA,AN,NA,AG,GA,AR$$
Now $$\quad P\left( TA/TATANAGAR \right) ={ \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) }/{ \left( \begin{matrix} 8 \\ 1 \end{matrix} \right) }={ 2 }/{ 8 }$$
and $$ P\left( TA/CALCUTTA \right) ={ 1 }/{ \left( \begin{matrix} 7 \\ 1 \end{matrix} \right) }={ 1 }/{ 7}$$
since both CALCUTTA and TATANAGAR are equally likely
$$ \Rightarrow P\left( TA \right) ={ 1 }/{ 2 }\times \left\{ { 2 }/{ 8 }+{ 1 }/{ 7 } \right\} ={ 11 }/{ 56 }$$
Hence by Bayes' theorem :
$$P\left( CALCUTTA/TA \right) =\dfrac { P\left( TA/CALCUTTA \right) P\left( CALCUTTA \right) }{ P\left( TA \right) } =\dfrac { 1 }{ 7 } \times \dfrac { 1 }{ 2 } \times \dfrac { 56 }{ 11 } =\dfrac { 4 }{ 11 }$$
Question 7
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A bag contains some white and some black balls, all of which are distinguishable from each other, all combinations of balls being equally likely. The total number of balls in the bag is $$10$$. If three balls are drawn at random and all of them are found to be black, the probability that the bag contains $$1$$ white and $$9$$ black balls is :

A
$$\displaystyle \frac{14}{55}$$

B
$$\displaystyle \frac{12}{55}$$

C
$$\displaystyle \frac{8}{55}$$

D
$$\displaystyle \frac{2}{11}$$

Solution

Possible cases are:
(i) $$3B, 7W$$; (ii) $$4B, 6W$$; (iii) $$5B, 5W$$; (iv) $$6B, 4W$$;
(v) $$7B, 3W$$; (vi) $$8B, 2W$$; (vii) $$9B, 1W$$; (viii) $$10B, 0W$$
Probability of each case $$=\dfrac { 1 }{ 8 } $$
Probability of getting $$3B$$ in the above cases:
(i) $$\dfrac { ^{ 3 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (ii) $$\dfrac { ^{ 4 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (iii) $$\dfrac { ^{ 5 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (iv) $$\dfrac { ^{ 6 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$

(v) $$\dfrac { ^{ 7 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (vi) $$\dfrac { ^{ 8 }C_{ 3 } }{ ^{ 10 }C_{ 3 } }$$; (vii) $$\dfrac { ^{ 9 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$; (viii) $$\dfrac { ^{ 10 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } $$

Therefore, required probability is
$$ \displaystyle \dfrac { \dfrac { ^{ 9 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } \times \dfrac { 1 }{ 8 } }{ \left( \dfrac { ^{ 3 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 4 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 5 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 6 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 7 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 8 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 9 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } +\dfrac { ^{ 10 }C_{ 3 } }{ ^{ 10 }C_{ 3 } } \right) \dfrac { 1 }{ 8 } } =\dfrac { 84 }{ 330 } =\dfrac { 14 }{ 55 } $$

Hence, A is correct.
Question 8
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A man is known to speak the truth $$3$$ out of $$4$$ times. He throws a die and reports that it is a six. The probability that it is actually a six is

A
$$\displaystyle \frac{3}{8}$$

B
$$\displaystyle \frac{1}{8}$$

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

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

Solution

$$\textbf{Step1: Find the probability that it is actually a six }$$
$$\text{Given:-}$$
  $$\text{Speaks the truth $$3$$ out of $$4$$ times}$$
  $$\text{Throws a die and reports that it is six}$$
  $$\text{Probability that the man speaks the truth is}$$ $$ = P(A) =\dfrac{3}{4}$$
  $$\text{The probability that the man lies is = }$$ $$P(B) = 1-{P(A) = }\bigg(1-\dfrac{3}{4}\bigg)=\dfrac{1}{4}$$
  $$\text{probability of getting}$$ $$6=\dfrac{1}{6}$$
  $$\text{probability of not getting}$$ $$6=\dfrac{5}{6}$$

  $$P(A|B)=\dfrac {P(B|A).P(A)}{P(B)}$$

  $$\text{where, }$$$$P(A|B)=$$ $$\text{probability of A given B is true}$$
  $$P(B|A)=$$ $$\text{probability of B given A is true}$$
  $$P(A), P(B)=$$ $$\text{the independent probabilities of A and B}$$

  $$\text{Applying Bayes' theorem, we get the required probability }$$$$=\dfrac{\dfrac{1}{6}.\dfrac{3}{4}}{\dfrac{1}{6}.\dfrac{3}{4}+\dfrac{5}{6}.\dfrac{1}{4}}$$

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

$$\textbf{Hence, the probability that it is actually a six is }$$ $$=\dfrac{3}{8}$$
Question 9
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5 cards are drawn at random from a well shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that there are 4 hearts is

A
$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}+^{13}C_{4}+^{13}C_{5}}$$

B
$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{39}C_{2}+^{13}c_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

C
$$\displaystyle \frac{^{13}C_{4}\times ^{39}C_{1}}{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

D
$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{13}C_{4}\times ^{13}C_{5}}$$

Solution

The cases possible are:
(i) 3 H, 2 non-H: $$^{ 13 }{ C_3 }*^{ 39 }{ C_2 }$$
(ii) 4 H, 1 non-H: $$^{ 13 }{ C_4 }*^{ 39 }{ C_1 }$$
(iii) 5H: $$^{ 13 }{ C_5 }$$
Hence, probability = $$\dfrac { ^{ 13 }{ C_4 }*^{ 39 }{ C_1 } }{^{ 13 }{ C_3 }*^{ 39 }{ C_2 }+^{ 13 }{ C_4 }*^{ 39 }{ C_1 }+^{ 13 }{ C_5 } } $$
Question 10
1 / -0

5 cards are drawn at random from a well shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that there are exactly 3 hearts is

A
$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}+^{13}c_{4}+^{13}c_{5}}$$

B
$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

C
$$\displaystyle \frac{^{13}c_{3}\times ^{39}c_{2}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

D
$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{13}c_{4}\times ^{13}c_{5}}$$

Solution

The cases possible are:
(i) 3 H, 2 non-H: $$_{ 3 }^{ 13 }{ C }*_{ 2 }^{ 39 }{ C }$$
(ii) 4 H, 1 non-H: $$_{ 4 }^{ 13 }{ C }*_{ 1 }^{ 39 }{ C }$$
(iii) 5H: $$_{ 5 }^{ 13 }{ C }$$
Hence, probability = $$\dfrac { _{ 3 }^{ 13 }{ C }*_{ 2 }^{ 39 }{ C } }{ _{ 3 }^{ 13 }{ C }*_{ 2 }^{ 39 }{ C }+_{ 4 }^{ 13 }{ C }*_{ 1 }^{ 39 }{ C }+_{ 5 }^{ 13 }{ C } } $$
Hence, (c) is correct.

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Re-Attempt Weekly Quiz Competition

Probability Test - 46

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Probability Test - 46

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SHARING IS CARING

Question 1

$${ \left( \cfrac { 1 }{ 2 } \right) }^{ 5 }$$

$${10}^{-1}$$

$$\cfrac{9}{10}$$

$${ \left( \cfrac { 9 }{ 10 } \right) }^{ 5 }$$

Solution

Question 2

$$\dfrac{1}{19}$$

$$\dfrac{2}{19}$$

$$\dfrac{3}{19}$$

$$\dfrac{1}{18}$$

Solution

Question 3

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

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

$$\displaystyle \frac{3}{9}$$

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

Solution

Question 4

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

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

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

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

Solution

Question 5

$$\displaystyle \frac{11}{1900}$$

$$\displaystyle \frac{44}{1900}$$

$$\displaystyle \frac{66}{1900}$$

$$\displaystyle \frac{99}{1900}$$

Solution

Question 6

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

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

$$\displaystyle \frac{1}{22}$$

$$\displaystyle \frac{21}{22}$$

Solution

Question 7

$$\displaystyle \frac{14}{55}$$

$$\displaystyle \frac{12}{55}$$

$$\displaystyle \frac{8}{55}$$

$$\displaystyle \frac{2}{11}$$

Solution

Question 8

$$\displaystyle \frac{3}{8}$$

$$\displaystyle \frac{1}{8}$$

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

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

Solution

Question 9

$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}+^{13}C_{4}+^{13}C_{5}}$$

$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{39}C_{2}+^{13}c_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

$$\displaystyle \frac{^{13}C_{4}\times ^{39}C_{1}}{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

$$\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{13}C_{4}\times ^{13}C_{5}}$$

Solution

Question 10

$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}+^{13}c_{4}+^{13}c_{5}}$$

$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

$$\displaystyle \frac{^{13}c_{3}\times ^{39}c_{2}}{^{13}c_{3}\times ^{39}c_{2}+^{13}c_{4}\times ^{39}c_{1}+^{13}c_{5}}$$

$$\displaystyle \frac{^{13}c_{4}}{^{13}c_{3}\times ^{13}c_{4}\times ^{13}c_{5}}$$

Solution

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