Probability Test

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Probability Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Find the chance that a non-leap year contains $$53$$ Saturdays

A
$$\dfrac {2}{7}$$

B
$$\dfrac {1}{7}$$

C
$$\dfrac {2}{365}$$

D
$$\dfrac {1}{365}$$

Solution

non-leap year has $$365$$ days which is divided into $$7 \times 52=364 + 1$$ days. There will be $$52$$ fridays, one for each week except when the first day of the year is a Friday. That probability is $$P(E)=\dfrac{1}{7}$$
Question 2
1 / -0

A card is drawn from a well shuffled deck of $$52$$ cards. Find the probability of getting a club card?

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

B
$$\dfrac {3}{4}$$

C
$$\dfrac {1}{4}$$

D
$$1$$

Solution

a well shuffled deck of 52 cards there are 13 club card
Hence $$P(E)=\dfrac{13}{52}=\dfrac{1}{4}$$
Question 3
1 / -0

Observe the table given.
Club English Maths Science
Number of students $$99$$ $$121$$ $$154$$
A student is chosen randomly from the group as the co-ordinator of all the clubs. What is the probability that the student is from the science club?

A
$$\dfrac {7}{17}$$

B
$$\dfrac {9}{34}$$

C
$$\dfrac {10}{17}$$

D
$$\dfrac {11}{34}$$

Solution

A student is chosen randomly from the group as the coordinator of all the clubs the probability that the student is from the science club

$$P(E)=\dfrac{student_science}{total_students}=\dfrac{154}{154+99+121}=\dfrac{7}{17}$$
Question 4
1 / -0

From a normal pack of cards, a card is drawn at random. Find the probability of getting a jack or a king.

A
$$\dfrac {2}{52}$$

B
$$\dfrac {1}{52}$$

C
$$\dfrac {2}{13}$$

D
$$\dfrac {1}{26}$$

Solution

From a normal pack of cards, a card is drawn at random
There are 4 kings and 4 jacks in a standard $$52$$ card deck. So the probability of drawing a king or a jack would be $$P(E)=\dfrac{8}{52}=\dfrac{2}{13}$$
Question 5
1 / -0

A bag contains a number of marbles of which $$80$$ are red, $$24$$ are white and the rest are blue. If the probability of randomly selecting a blue marble from this bag is $$\dfrac {1}{5}$$, how many blue marbles are there in the bag?

A
$$25$$

B
$$26$$

C
$$27$$

D
$$28$$

Solution

A bag contains a number of marbles of which 80 are red, 24 are white and the rest are blue. If the probability of randomly selecting a blue marble from this bag is $$\dfrac{1}{5}$$
The Probabaility of red marble is $$P(E)=\dfrac{1}{5}=$$$$\dfrac{blue}{red+blue+white}$$ = $$\dfrac{blue}{80+24+blue}$$
from Above we get
$$4$$ blue $$=104$$
blue $$=26$$
Question 6
1 / -0

A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing a face card.

A
$$\dfrac { 8 }{ 13 } $$

B
$$\dfrac { 3 }{ 13 } $$

C
$$\dfrac { 4 }{ 13 } $$

D
$$\dfrac { 1 }{ 4 } $$

Solution

There are $$12$$ face cards in 52 well-shuffled deck of playing cards
therefore $$P(E)=\dfrac{card_{face}}{card_{total}}=$$$$\dfrac{12}{52}=$$$$\dfrac{3}{13}$$
Question 7
1 / -0

A card is drawn at random from a pack of well shuffled $$52$$ cards. What is the probability of getting a queen of red suit?

A
$$\dfrac { 1 }{ 26 } $$

B
$$\dfrac { 1 }{ 13 } $$

C
$$\dfrac { 1 }{ 52 } $$

D
$$\dfrac { 1 }{ 14 } $$

Solution

there are $$2$$ queen of red suit.
A card is drawn at random from a pack of well shuffled 52 cards.

$$P(E)=\dfrac{2}{52}=\dfrac{1}{26}$$
Question 8
1 / -0

In a single throw of two dice, find the probability that neither a doublet nor a total of 8 will appear.

A
$$\cfrac{11}{36}$$

B
$$\cfrac{5}{18}$$

C
$$\cfrac{13}{18}$$

D
$$\cfrac{3}{16}$$

Solution

Doublet = $$\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$$

Total $$8$$ will be from the combination of $$\{(2,6),(6,2),(3,5),(5,3),(4,4)\}$$

Total no. of combinations $$6+5-1=10 $$ ($$-1$$ because $$(4, 4)$$ taken two times)

So, no. of possibilities that neither a doublet nor a total of 8 will be $$=36-10=26 $$

Total no. possibilities $$=6\times 6=36$$

So, Probability $$=\dfrac{26}{36}=\dfrac{13}{18}$$
Question 9
1 / -0

A ball is drawn at random from a box containing 10 red, 30 white, 20 blue and 15 orange marbles. The probability of a ball drawn is red, white or blue .......

A
$$\dfrac 13$$

B
$$\dfrac 35$$

C
$$\dfrac 23$$

D
$$\dfrac 45$$

Solution

Total number of balls in the box, $$n(S)=10+30+20+15=75$$
Number of red balls, 10
Number of white balls, 30
Number of blue balls, 20
Hence the number of favourable outcomes for getting a red, white or blue ball is, $$n(E)=10+30+20=60$$
Hence the probability of a ball drawn is red, white or blue is, $$ \dfrac {n(E)}{n(S)}=\dfrac {60}{75}=\dfrac 45$$
Question 10
1 / -0

In a non-leap year the probability of getting $$53$$ Sundays or $$53$$ Tuesdays or $$53$$ Thursdays is.

A
$$\dfrac {1}{7}$$

B
$$\dfrac {2}{7}$$

C
$$\dfrac {3}{7}$$

D
$$\dfrac {4}{7}$$

Solution

Non-leap year has $$365$$ days

$$1$$ year $$=52$$ weeks

$$=52\times 7 days$$

$$=364$$ days [It contains $$52$$ Sundays, Mondays and so on].

$$\therefore 1$$ day will be left sample space for this

$$1$$ day $$=\{$$Mon, Tue, Wed, Thurs, Fri, Sat, Sun$$\} $$

$$P[ 53$$ Sundays$$] =\dfrac { 1 }{ 7 } $$

$$P[ 53$$ Tuesdays$$] =\dfrac { 1 }{ 7 } $$

$$P[53$$ Thursdays$$] =\dfrac { 1 }{ 7 } $$

$$P(53$$ Sundays or $$53$$ Tuesdays or $$53$$ Thursdays$$)$$

$$=P(53$$ Sundays$$)+P(53$$ Tuesdays$$)+P(53$$ Thursdays$$)$$

$$=\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 7 } $$

$$={ \dfrac { 3 }{ 7 } } $$

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Re-Attempt Weekly Quiz Competition

Club	English	Maths	Science
Number of students	$$99$$	$$121$$	$$154$$

Probability Test - 18

Result

Probability Test - 18

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SHARING IS CARING

Question 1

$$\dfrac {2}{7}$$

$$\dfrac {1}{7}$$

$$\dfrac {2}{365}$$

$$\dfrac {1}{365}$$

Solution

Question 2

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

$$\dfrac {3}{4}$$

$$\dfrac {1}{4}$$

$$1$$

Solution

Question 3

$$\dfrac {7}{17}$$

$$\dfrac {9}{34}$$

$$\dfrac {10}{17}$$

$$\dfrac {11}{34}$$

Solution

Question 4

$$\dfrac {2}{52}$$

$$\dfrac {1}{52}$$

$$\dfrac {2}{13}$$

$$\dfrac {1}{26}$$

Solution

Question 5

$$25$$

$$26$$

$$27$$

$$28$$

Solution

Question 6

$$\dfrac { 8 }{ 13 } $$

$$\dfrac { 3 }{ 13 } $$

$$\dfrac { 4 }{ 13 } $$

$$\dfrac { 1 }{ 4 } $$

Solution

Question 7

$$\dfrac { 1 }{ 26 } $$

$$\dfrac { 1 }{ 13 } $$

$$\dfrac { 1 }{ 52 } $$

$$\dfrac { 1 }{ 14 } $$

Solution

Question 8

$$\cfrac{11}{36}$$

$$\cfrac{5}{18}$$

$$\cfrac{13}{18}$$

$$\cfrac{3}{16}$$

Solution

Question 9

$$\dfrac 13$$

$$\dfrac 35$$

$$\dfrac 23$$

$$\dfrac 45$$

Solution

Question 10

$$\dfrac {1}{7}$$

$$\dfrac {2}{7}$$

$$\dfrac {3}{7}$$

$$\dfrac {4}{7}$$

Solution

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