Probability Test

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Probability Test - 33

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Weekly Quiz Competition

Question 1
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A coin is tossed $$40$$ times and it showed tail $$24$$ times.The probability of getting a head was:

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

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

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

D
$$\displaystyle\frac{17}{40}$$

Solution

Coin tossed $$=40$$ times
showed tail $$=24$$ times
Probability (getting head) $$=\dfrac{40-24}{40}$$
$$=\dfrac{16}{40}$$
$$=\dfrac{2}{5}$$
Question 2
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Directions For Questions

In a multiple choice test of three questions there are five alternative answers given to the first two questions each and four alternative answers given to the last question. If a candidate guesses answers at random, what is the probability that he will get-

...view full instructions

Exactly one right?

A
$$\dfrac {2}{5}$$

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

C
$$\dfrac {3}{5}$$

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

Solution

Given,In a multiple choice test there are five alternate options for first two question and 4 alternate options for third question.
The probability to do first question correctly,$$P(F1C)=\dfrac{1}{5}$$

The probability to do second question correctly,$$P(F2C)=\dfrac{1}{5}$$

The probability to do third question correctly,$$P(F3C)=\dfrac{1}{4}$$

Similarly,The probability to do first question wrong,$$P(F1U)=\dfrac{4}{5}$$

The probability to do second question wrong,$$P(F2U)=\dfrac{4}{5}$$

The probability to do third question wrong,$$P(F3U)=\dfrac{3}{4}$$

The probability to do exactly one question correctly=(Probability to do first question correct and rest wrong)+(Probability to do second question correct and rest wrong)+(Probability to do third question correct and rest wrong)
$$=(P(F1C)*P(F2U)*P(F3U))+(P(F2C)*P(F1U)*P(F3U))+(P(F3C)*P(F2U)*P(F1U))$$
$$=(\displaystyle\frac{1}{5}*\frac{4}{5}*\frac{3}{4})+(\frac{1}{5}*\frac{4}{5}*\frac{3}{4})+(\frac{1}{4}*\frac{4}{5}*\frac{4}{5})$$

$$P=\displaystyle\frac{(3+3+4)}{25}=\frac{2}{5}$$
Question 3
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One ticket is selected at random from $$100$$ tickets numbered $$00,01,02,...,99$$. Suppose $$X$$ and $$Y$$ are the sum and product of the digits found on the tickets $$P(X=7;Y=0)$$ is given by

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

B
$$\displaystyle \frac {2}{19}$$

C
$$\displaystyle \frac {1}{50}$$

D
None of these

Solution

We have $$\left( X=7 \right) =\left\{ 07,16,25,34,43,52,61,70 \right\} $$
and $$\left( Y=0 \right) =\left\{ 00,01,02,...,10,20,30,..,90 \right\} $$
Thus $$\left( X=7 \right) \cap \left( Y=0 \right) =\left\{ 07,70 \right\} $$
$$\displaystyle \therefore P\left( X=7;Y=0 \right) =\frac { 2 }{ 100 } =\frac { 1 }{ 50 } $$
Question 4
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In a class of 125 students, 70 passed in English, 55 in mathematics and 30 in both. Find the probability that a student selected at random from the class has passed in at least one subject.

A
$$\displaystyle \frac {19}{25}$$

B
$$\displaystyle \frac {6}{25}$$

C
$$\displaystyle \frac {17}{25}$$

D
$$\displaystyle \frac {8}{25}$$

Solution

In a class of $$125$$ students, $$70$$ passed in English, $$55$$ passed in mathematics and $$30$$ passed in both.

$$\Rightarrow n(\mu)=125, n(E)=70, n(M)=55, n(E \cap M)=30$$

Total no.of students passed $$=n(E \cup M)=n(E)+n(M)-n(M \cap E)=70+55-30=95$$

$$\therefore$$ The total no.of students failed $$=n(U)-n(E \cup M)=125-95=30$$

The probability that student passed in atleast one subject $$=\dfrac{n(E \cup M)}{n(U)}$$

$$=\dfrac{95}{125}=\dfrac{19}{25}$$
Question 5
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If E & F are events with $$P ( E) \leq P (F)$$ & $$P(E \cap F ) > 0$$, then

A
occurrence of E $$\Rightarrow$$ occurrence of F

B
occurrence of F$$\Rightarrow$$ occurrence of E

C
non occurrence of E $$\Rightarrow$$ non occurrence of F

D
none of the above implications holds

Solution

Given,$$P(E)\leq P(F)\;and\;P(E \cap F)>0$$
$$\Rightarrow$$ there is a some common exist between E and F which is shown in below image.
1) Occurence of E doesn't always $$\Rightarrow$$ occurence of F,since $$E\neq F$$ always.when E<F,There is chance of occuring E irrespective of occurence of F
2)Similary Occurence of F doesn't always $$\Rightarrow$$ occurence of E,since $$E\neq F$$.when E<F,There is chance of occuring F irrespective of occurence of E
3)Similary Occurence of E doesn't always $$\Rightarrow$$ non-occurence of F,since $$E\neq F$$.There might be a chance of F to occur irrespective of E not occuring.
$$\therefore $$ None of the given options are correct.
Question 6
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Three numbers are chosen at random without replacement from $${1, 2, 3,....., 10}$$. The probability that the minimum of the chosen numbers is $$3$$ or their maximum is $$7$$ is

A
$$1/2$$

B
$$1/3$$

C
$$1/4$$

D
$$11/40$$

Solution

Selection of 3 no.'s from {1,2,3,4,5,6,7,8,9,10}
$$\Rightarrow$$ No.of combinations to be made=$$^{10}C_{3}$$
Selection of 3 no.'s having 3 as minimum should be selected from the set {3,4,5,6,7,8,9,10} for which one number is fixed i.e.,3
$$\Rightarrow$$ No.of combinations to be made=$$^{7}C_{2}$$
selection of 3 no.'s having 7 as maximum should be selected from the set {1,2,3,4,5,6,7} for which one number is fixed i.e.,7
$$\Rightarrow$$ No.of combinations to be made=$$^{6}C_{2}$$
selection of 3 no.'s having 7 as maximum and 3 as minimum should be selected from the set {3,4,5,6,7} for which two numbers fixed i.e.,3 and 7
$$\Rightarrow$$ No.of combinations to be made=$$^{3}C_{1}$$
Therefore the probability that the minimum of the chosen numbers is 3 or their maximum is 7 is=$$(\dfrac{^{7}C_{2}+^{6}C_{2}-^{3}C_{1}}{^{10}C_{3}})$$
=$$\displaystyle\dfrac{(21+15-3)}{120}$$=$$\dfrac{33}{120}=\dfrac{11}{40}$$
Question 7
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In a horse race there are 18 horses numbered from 1 to 18. The probability that horse 1 would win is $$\displaystyle \frac {1}{6}$$, horse 2 is $$\displaystyle \frac {1}{10}$$ and 3 is $$\displaystyle \frac {1}{8}$$. Assuming a tie is impossible, the chance that one of the three horses wins the race, is

A
$$\displaystyle \frac {143}{420}$$

B
$$\displaystyle \frac {119}{120}$$

C
$$\displaystyle \frac {47}{120}$$

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

Solution

Give,18 horses are numbered as 1,2,3,...,18 respectively.
The probability that horse1 would win the race,$$P(H1)=\dfrac{1}{6}$$
The probability that horse2 would win the race,$$P(H2)=\dfrac{1}{10}$$
The probability that horse3 would win the race,$$P(H3)=\dfrac{1}{8}$$
The probability that one of the horse1 or horse2 or horse3 win the race$$=P(H1)+P(H2)+P(H3)$$
$$\therefore The\;required\;probability=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{8}=\dfrac{47}{120}$$
Question 8
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Three unbiased coins are tossed, what is probability of getting exactly two heads ?

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

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

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

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

Solution

Possible outcomes of tossing three coins are:
$$ (HHH), (HHT), (HTH), (THH), (TTT), (TTH), (THT),(HTT) $$
here $$H $$ and $$ T$$ are denoted for Head and Tail.
Total no. of outcomes $$= 8$$
no. of outcomes with exactly two heads $$=$$$$ 3$$
$$\therefore $$ required probability = $$ \dfrac 38 $$
$$\therefore $$ Option D is correct.
Question 9
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All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting black face card

A
$$\displaystyle \frac{6}{49}$$

B
$$\displaystyle \frac{3}{49}$$

C
$$\displaystyle \frac{5}{49}$$

D
$$\displaystyle \frac{4}{49}$$

Solution

Total number of possibilities = $$49$$ (Since, 3 cards of spade are removed)
Number of black face cards = $$3$$ (3 cards of clubs)
Thus, required probability = $$\dfrac{3}{49}$$
Question 10
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Directions For Questions

A uniform unbiased die is constructed in the shape of a regular tetrahedron with faces numbered $$2, 2, 3$$ and $$4$$ and the score is taken from the face on which the die lands. If two such dice are thrown together, find the probability of scoring.

...view full instructions

Exactly 6 on each of 3 successive throws

A
$$\displaystyle \frac {5^3}{16^3}$$

B
$$\displaystyle \frac {5}{16^3}$$

C
$$\displaystyle \frac {5^2}{16^3}$$

D
$$\displaystyle \frac {1}{16^3}$$

Solution

Total = 6
Possible combinations $$\{2,4\}\{3,3\}$$.
$$P(2,4) = \cfrac{2}{4} \cfrac{1}{4}.2 = 1/4$$
$$P(3,3) = \cfrac{1}{4} \cfrac{1}{4} = 1/16$$
$$P(sum = 6) = P(2,4)+P(3,3)= 5/16$$
P(Exactly 6 on each of 3 successive throws) = $$P(sum = 6)^3 = (5/16)^3$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 33

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Probability Test - 33

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SHARING IS CARING

Question 1

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

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

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

$$\displaystyle\frac{17}{40}$$

Solution

Question 2

$$\dfrac {2}{5}$$

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

$$\dfrac {3}{5}$$

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

Solution

Question 3

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

$$\displaystyle \frac {2}{19}$$

$$\displaystyle \frac {1}{50}$$

None of these

Solution

Question 4

$$\displaystyle \frac {19}{25}$$

$$\displaystyle \frac {6}{25}$$

$$\displaystyle \frac {17}{25}$$

$$\displaystyle \frac {8}{25}$$

Solution

Question 5

occurrence of E $$\Rightarrow$$ occurrence of F

occurrence of F$$\Rightarrow$$ occurrence of E

non occurrence of E $$\Rightarrow$$ non occurrence of F

none of the above implications holds

Solution

Question 6

$$1/2$$

$$1/3$$

$$1/4$$

$$11/40$$

Solution

Question 7

$$\displaystyle \frac {143}{420}$$

$$\displaystyle \frac {119}{120}$$

$$\displaystyle \frac {47}{120}$$

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$\displaystyle \frac{6}{49}$$

$$\displaystyle \frac{3}{49}$$

$$\displaystyle \frac{5}{49}$$

$$\displaystyle \frac{4}{49}$$

Solution

Question 10

$$\displaystyle \frac {5^3}{16^3}$$

$$\displaystyle \frac {5}{16^3}$$

$$\displaystyle \frac {5^2}{16^3}$$

$$\displaystyle \frac {1}{16^3}$$

Solution

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