Probability Test

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Probability Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

A card is drawn at random from well shuffled pack of $$52$$ cards. Find the probability that the card drawn is not of diamond:

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

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

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

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

Solution

Sample space $$= 52$$
Event a card obtained is not diamond $$= 13$$(spade)+$$13$$(club)+$$13$$(hearts)=$$39$$
Probability$$=\cfrac{39}{52}$$ $$=\cfrac{3}{4}$$
Question 2
1 / -0

If three coins are tossed then find the probability of the events of getting exactly one tail.

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

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

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

D
$$\displaystyle \frac{5}{8}$$

Solution

A coin tossed three times sample space $$= {(H,H,H),(H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H),(T,T,T)}$$
favourable outcomes for Exactly on tail $$= 3$$
Probability = $$\cfrac{3}{8}$$
Question 3
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If three coins are tossed then find the probability of the event of getting no tail.

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

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

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

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

Solution

A coin tossed three times sample space = {(H,H,H),(H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H),(T,T,T)}
Favourable outcomes for No tail = $$1$$
Probability = $$\cfrac{1}{8}$$
Question 4
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A box contains $$20$$ cards marked with the numbers $$ 1 $$ to $$20$$. One card is drawn from this box at random. What is the probability that number on the card is a multiple of $$5$$?

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

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

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

D
$$\displaystyle \frac{4}{5}$$

Solution

Sample space $$= 20$$
multiples of 5 are$$ (5,10,15,20)$$
probability of perfect square $$=\cfrac{4}{20}$$$$=\cfrac{1}{5}$$
Question 5
1 / -0

A card is drawn at random from well shuffled pack of $$52$$ cards. Find the probability that the card drawn is a spade:

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

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

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

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

Solution

Sample space $$= 52$$
Event a card obtained is of spade $$= 13$$(spades ,$$13$$ cards of each suits)
Probability $$=\cfrac{13}{52}$$ $$=\cfrac{1}{4}$$
Question 6
1 / -0

Two die are thrown. Find the probability that the number on the upper face of the first dice is less than the number on the upper face of the second dice.

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

B
$$\displaystyle \frac{7}{12}$$

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

D
$$\displaystyle \frac{5}{12}$$

Solution

Sample space of nos obtained is $$= 6\times6 $$ (As two dices are independent)
No. obtained on first dice is less than second $$= 15 $$
(if no on first dice is $$1$$ then possibility on second dice are $$ 5 (2-6)$$ and similarly for $$2= 4$$ , for $$ 3= 3 $$, for $$4 =2$$ ,for $$5 =1$$ total $$5+ 4 +3+2+1=15)$$
probability of event $$= 15/36= 5/12$$
Question 7
1 / -0

Two die are thrown find the probability of getting the sum of the numbers on their upper faces to be at most $$3$$.

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

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

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

D
$$\displaystyle \frac{1}{36}$$

Solution

Sample space of nos obtained is $$= 6\times6 $$ = $$36$$ (As two dices are independent)

Sum of nos. obtained atmost $$ 3 = {(1,1),(1,2)(2,1)}=3$$
probability of getting the event = $$\cfrac{3}{36}$$ =$$\cfrac{1}{12}$$
Question 8
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Two fair die are thrown, find the probability that sum of the points on their uppermost faces is a perfect square or divisible by $$4$$:

A
$$\displaystyle \frac{12}{36}$$

B
$$\displaystyle \frac{13}{36}$$

C
$$\displaystyle \frac{15}{36}$$

D
$$\displaystyle \frac{16}{36}$$

Solution

two fair dice thrown
so total cases$$= 6\times6=36$$
and perfect square or divisible by $$4$$ that mean that sum should be equal to $$=4,8,9,12$$
for getting $$4= (1,3);(2,2);(3,1)$$
for getting $$8= (2,6);(3,5);(4,4);(5,3);(6,2)$$
for getting $$ 9= (3,6);(4,5);(5,4);(6,3)$$
for getting $$12=(6,6)$$
so total cases$$=3+5+4+1=13$$
so probability =$$\cfrac{13}{36}$$
Question 9
1 / -0

Two die are thrown find the probability of getting the sum of the numbers on their upper faces divisible by 9.

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

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

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

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

Solution

Sample space of nos obtained is $$= 6\times6 $$ (As two dices are independent)
Sum of nos. obtained divisible by $$9 = {(3,6),(4,5)(5,4),(6,3)}=4$$
probability of getting the event = $$\cfrac{4}{36}$$= $$\cfrac{1}{9}$$
Question 10
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 $$=\dfrac{3}{9} = \dfrac{1}{3}$$

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Probability Test - 21

Result

Probability Test - 21

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

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

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

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

$$\displaystyle \frac{5}{8}$$

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

$$\displaystyle \frac{4}{5}$$

Solution

Question 5

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

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

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

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

Solution

Question 6

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

$$\displaystyle \frac{7}{12}$$

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

$$\displaystyle \frac{5}{12}$$

Solution

Question 7

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

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

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

$$\displaystyle \frac{1}{36}$$

Solution

Question 8

$$\displaystyle \frac{12}{36}$$

$$\displaystyle \frac{13}{36}$$

$$\displaystyle \frac{15}{36}$$

$$\displaystyle \frac{16}{36}$$

Solution

Question 9

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

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

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

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

Solution

Question 10

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

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

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

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

Solution

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