Probability Test

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Probability Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The probability of getting number less than or equal to $$6$$, when a die is thrown once, is

A
An impossible event

B
A sure event

C
An exhaustive event

D
A complementary event

Solution

The probability of getting number less than $$6$$, when a die is thrown once, is a sure event.
Because, once a die is thrown, sample space $$= \{1, 2, 3, 4, 5, 6\}$$
There is a possible event for getting number less than $$6$$ as outcomes can be $$1, 2, 3, 4, 5$$.
Question 2
1 / -0

The letters of the alphabet are written on $$26$$ cards. Two cards are chosen at random. Find the probability that at least one of them is a vowel?

A
$$\dfrac{23}{65}$$

B
$$\dfrac{13}{24}$$

C
$$\dfrac{5}{65}$$

D
$$\dfrac{1}{23}$$

Solution

The complement of $$A'$$ at least one of them is vowels is neither of them is a vowel.
There are $$5$$ vowels and $$21$$ are consonants $$P$$ (The first card chosen is not a vowel) $$=$$ $$\dfrac{21}{26}$$
$$P$$ (The second card chosen is also not a vowel) $$=$$ $$\dfrac{20}{25}$$
$$P$$ (Neither of them is a vowel) $$=$$ $$\dfrac{21}{26}\times \dfrac{20}{25}= \dfrac{42}{65}$$
$$P$$ (At least one of them is a vowel) $$=$$ $$1-\dfrac{42}{65} = \dfrac{23}{65}$$
Question 3
1 / -0

Two fair coins are tossed. What is the probability that at least one coin lands a tail?

A
$$\dfrac{1}{5}$$

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

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

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

Solution

$$P$$ (Both coins land heads) $$=$$ $$\dfrac{1}{2}\times \dfrac{1}{2} = \dfrac{1}{4}$$

$$P$$(At least one coin lands tails) $$=$$ $$1-\dfrac{1}{4} = \dfrac{3}{4}$$
Question 4
1 / -0

A coin is tossed and a single $$6$$-sided die is rolled. Find the probability of landing on the tail side of the coin and rolling $$4$$ on the die.

A
$$\dfrac{1}{12}$$

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

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

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

Solution

$$P$$ (tail) $$=$$ $$\dfrac{1}{2}$$ and $$P(4) =$$ $$\dfrac{1}{6}$$

$$P$$ (tail and $$4$$) $$=$$ $$P$$(tail) $$. P(4)$$
$$=$$$$\cfrac{1}{2}\times \cfrac{1}{6}$$ $$=$$ $$\cfrac{1}{12}$$
Question 5
1 / -0

From a standard deck of cards, one card is drawn. Find the probability that the card is red and a queen.

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

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

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

D
$$\dfrac{12}{13}$$

Solution

Total number of playing cards $$= 52$$
Number of red queen in a deck of $$52$$ cards $$= 2$$
$$P$$ (the card is red and a queen) $$ =$$ $$\dfrac{\text{Number space of space red queen card}}{\text{Total space number space of space playing space cards}}$$
$$=$$ $$\dfrac{2}{52}= \dfrac{1}{26}$$
Question 6
1 / -0

You roll a pair of dice. What is the probability of not getting a double on the dice?

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

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

C
$$\dfrac{2}{6}$$

D
$$\dfrac{7}{6}$$

Solution

There are $$6$$ ways to roll pair of dice.
$$P$$ (doubles) $$=$$ $$\dfrac{6}{36} = \dfrac{1}{6}$$$$P$$ (not doubles) $$=$$ $$1-\dfrac{6}{36} = \dfrac{5}{6}$$
Question 7
1 / -0

If we throw a dice, then the sample space, $$S = {1, 2, 3, 4, 5, 6}$$. Now the event of $$3$$ appearing on the dice is simple and given by

A
$$E = {2, 3}$$

B
$$E = {1, 2, 3}$$

C
$$E = {3}$$

D
$$E = {1, 3}$$

Solution

If we throw a dice, then the sample space, $$S = {1, 2, 3, 4, 5, 6}$$. Now the event of 3 appearing on the dice is simple and given by $$E = {3}$$
If there be only one element of the sample space in the set representing an event, then this event is called a simple or elementary event.
Question 8
1 / -0

Two fair dice are thrown. What is the probability that the two scores do not add to $$5$$?

A
$$\dfrac{7}{9}$$

B
$$\dfrac{5}{9}$$

C
$$\dfrac{8}{9}$$

D
$$\dfrac{1}{9}$$

Solution

There are $$36$$ possible ways two dice can land, and of these $$4$$ of them add to $$5$$ i.e.
$$1 + 4 = 5$$
$$2 + 3 = 5$$
$$3 + 2 = 5$$
$$4 + 1 = 5$$
$$P$$ (The two scores do add to $$5$$) $$=$$ $$\dfrac{4}{36}=\dfrac{1}{9}$$
$$P$$ (The two scores do not add to $$5$$) $$=$$ $$1-\dfrac{1}{9}=\dfrac{8}{9}$$
Question 9
1 / -0

A coin is tossed $$100$$ times with the following frequencies: Head : $$20$$. Find the probability for event having heads only.

A
$$0.2$$

B
$$0.5$$

C
$$0.65$$

D
$$1.5$$

Solution

Since the coin is tossed $$100$$ times, the total number of trials is $$100$$.
Let us call the events of getting a head and of getting a tail as $$X$$ and $$Y$$, respectively. Then, the number of times $$X$$ happens, i.e., the number of times a head come up, is $$20$$.
$$P(X) =$$ $$\dfrac{\text{Number space of space heads}}{\text{Total space number space of space trials}}$$
$$= \dfrac{20}{100}= \dfrac{1}{5} = 0.2$$
Question 10
1 / -0

X and Y are independent events. If the probability event X will occur is $$0.2$$ and the probability event Y will occur is $$0.9$$, calculate the probability that both X and Y will occur.

A
$$18$$%

B
$$20$$%

C
$$50$$%

D
$$72$$%

E
$$90$$%

Solution

Given that the Probability that $$X$$ will occur is $$0.2$$ and Probability that $$Y$$ will occur is $$0.9$$
$$\Rightarrow$$ $$P(X)=0.2$$ and $$P(Y)=0.9$$
Given that $$X$$ and $$Y$$ are independent
So Probability that both $$X$$ and $$Y$$ occurs is $$P(X) \times P(Y) = 0.2 \times 0.9 = 0.18$$
The percentage is $$0.18 \times 100 = 18$$ percent

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Probability Test - 37

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Probability Test - 37

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SHARING IS CARING

Question 1

An impossible event

A sure event

An exhaustive event

A complementary event

Solution

Question 2

$$\dfrac{23}{65}$$

$$\dfrac{13}{24}$$

$$\dfrac{5}{65}$$

$$\dfrac{1}{23}$$

Solution

Question 3

$$\dfrac{1}{5}$$

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

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

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

Solution

Question 4

$$\dfrac{1}{12}$$

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

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

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

Solution

Question 5

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

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

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

$$\dfrac{12}{13}$$

Solution

Question 6

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

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

$$\dfrac{2}{6}$$

$$\dfrac{7}{6}$$

Solution

Question 7

$$E = {2, 3}$$

$$E = {1, 2, 3}$$

$$E = {3}$$

$$E = {1, 3}$$

Solution

Question 8

$$\dfrac{7}{9}$$

$$\dfrac{5}{9}$$

$$\dfrac{8}{9}$$

$$\dfrac{1}{9}$$

Solution

Question 9

$$0.2$$

$$0.5$$

$$0.65$$

$$1.5$$

Solution

Question 10

$$18$$%

$$20$$%

$$50$$%

$$72$$%

$$90$$%

Solution

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