Probability Test

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Probability Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number between 1 to 15. What is the probability that it will point to an odd number.

A
5/9

B
1/9

C
3/19

D
8/15

Solution

odd numbers between 1 to 15 are 1, 3, 5, 7, 9, 11, 13,15
So P(getting an odd number is)= $$\cfrac{8}{15}$$
Question 2
1 / -0

Two dice are thrown simultaneously. Find the probability of getting a total of at least $$10$$.

A
$$\dfrac{5}{6}$$

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

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

D
None of these

Solution

Total number of possible cases $$= 36$$
Favourable cases of getting a total of at least $$10$$
$$= \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}$$
Total number of favourable cases $$= 6$$
P(Total of at least $$10$$) = $$\displaystyle \frac{6}{36} = \frac{1}{6}$$
Question 3
1 / -0

A bulb is taken out at random from a box of 600 electric bulbs that contains 12 defective bulbs. Then the probability of a non-defective bulb is

A
0.02

B
0.98

C
0.50

D
None

Solution

Total number of bulbs = 600
Total number of non-defective bulb = 600-12=588
P (non-defective bulbs) = $$\displaystyle \frac{588}{600}= 0.98$$
Question 4
1 / -0

A pair of dice is thrown once. The probability that the sum of the outcomes is less than 11 is:

A
29/36

B
7/36

C
11/12

D
1/6

Solution

There are a total of 36 outcomes when two dice are thrown. All the pairs are Favorable outcomes except (6,6), (6,5), and (5,6) whose sum is equal to or more than 11.
Therefore $$36-3 = 33$$ outcomes that are favorable.
Hence,
$$P(sum<11)$$ $$=$$ $$\displaystyle \frac{33}{36}=\frac{11}{12}$$
Question 5
1 / -0

If a card is drawn from a pack of cards. The probability of getting black ace is

A
$$\displaystyle \frac{1}{52}$$

B
$$\displaystyle \frac{1}{26}$$

C
$$\displaystyle \frac{1}{13}$$

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

Solution

Total number of cases $$= 52$$
Number of favourable cases (getting a black ace) $$= 2$$
Thus, Probability (getting a black ace) $$=$$ $$\cfrac{2}{52} = \cfrac{1}{26}$$
Question 6
1 / -0

Two dice are thrown simultaneously. Find the probability of getting an even number as the sum.

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

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

C
$$\dfrac{5}{6}$$

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

Solution

Total number of possible cases $$= 36$$
Favourable cases of getting even number as the sum
$$= \{(1, 1), (1,3), (1,5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3,5), (4, 2), (4,4), (4,6), (5,1), \ \ \ \ \\ \ \ \ \ \ \ (5, 3), (5, 5), (6, 2), (6,4),\ (6, 6)\}$$
Total number of favourable cases $$= 18$$
$$P $$(getting even number as the sum) $$=$$ $$\displaystyle \frac{18}{36} = \frac{1}{2}$$
Question 7
1 / -0

Two dice are thrown simultaneously. Find the probability of getting the sum as a prime number.

A
$$\dfrac{15}{36}$$

B
$$\dfrac{5}{6}$$

C
$$\dfrac{14}{37}$$

D
None of these

Solution

Total number of possible cases $$= 36$$
Favourable cases of getting sum as a prime number
$$= \{(1, 1), (1,2), (1,4), (1,6), (2, 1), (2, 3), (2, 5), (3,2),(3,4), (4, 1), (4, 3), (5,2), (5, 6), (6,1), (6, 5)\}$$
Total number of favourable cases $$= 15$$
P (Getting a prime number) = $$\displaystyle \frac{15}{36}$$
Question 8
1 / -0

Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on first dice and a multiple of 3 on the second dice.

A
$$\displaystyle \frac{4}{6}$$

B
$$\displaystyle \frac{2}{6}$$

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

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

Solution

Total cases = $$6\times 6$$
Let A be the event of getting a multiple of 2 on first dice and a multiple of 3 on the second dice.
Hence, $$A= \{(2,3),(2,6),(4,3),(4,6),(6,3),(6,6)\}$$
$$n(A) =$$ $$6$$
$$\therefore $$ $$P(A) =$$ $$ \dfrac 6{36} = \dfrac 16 $$
$$\therefore $$ Option C is correct.
Question 9
1 / -0

A fair dice has faces numbered $$0, 1, 7, 3, 5$$ and $$9$$. If it is thrown, the probability of getting an odd number is:

A
$$1$$

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

C
$$\dfrac {5}{6}$$

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

Solution

A fair dice has faces numbered as $$0,1,7,3,5,9$$
Since $$1, 3, 5,7, 9$$ are odd numbers.
Number of favorable outcomes $$= 5$$
Total number of possible outcomes $$= 6$$
So, the probability of getting an odd number $$P(E)$$ $$=$$ $$\dfrac {\text{Favourable outcomes}}{\text{Total number of possible outcomes}}$$ $$=$$ $$\dfrac {5}{6}$$
Question 10
1 / -0

If two coins are tossed then find the probability of the event of getting one head .

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

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

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

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

Solution

sample space is ={h,t} ,{t,h}
favourable outcomes for getting one head =$$2 $$
probability (one head)= $$\cfrac{2}{4}$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 20

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Probability Test - 20

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SHARING IS CARING

Question 1

5/9

1/9

3/19

8/15

Solution

Question 2

$$\dfrac{5}{6}$$

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

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

None of these

Solution

Question 3

0.02

0.98

0.50

None

Solution

Question 4

29/36

7/36

11/12

1/6

Solution

Question 5

$$\displaystyle \frac{1}{52}$$

$$\displaystyle \frac{1}{26}$$

$$\displaystyle \frac{1}{13}$$

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

Solution

Question 6

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

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

$$\dfrac{5}{6}$$

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

Solution

Question 7

$$\dfrac{15}{36}$$

$$\dfrac{5}{6}$$

$$\dfrac{14}{37}$$

None of these

Solution

Question 8

$$\displaystyle \frac{4}{6}$$

$$\displaystyle \frac{2}{6}$$

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

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

Solution

Question 9

$$1$$

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

$$\dfrac {5}{6}$$

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

Solution

Question 10

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

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

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

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

Solution

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