Probability Test

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Probability Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

$$90\%$$ of the mangoes in a bag are good. If a mango is chosen randomly from the box, find the probability of getting a bad mango.

A
$$\dfrac{9}{100}$$

B
$$\dfrac{1}{100}$$

C
$$\dfrac{9}{10}$$

D
$$\dfrac{1}{10}$$

Solution

90% of the mangoes in a bag are good.
The number of bad mango are $$N=100-90=10\%$$
Therefore the probability of getting a bad mango $$P(E)=\dfrac{N}{100}=\dfrac{10}{100}=\dfrac{1}{10}$$
Question 2
1 / -0

Two dice are thrown. What is the probability that the sum of the two dice is greater than $$3$$?

A
$$1/4$$

B
$$3/4$$

C
$$5/6$$

D
$$11/12$$

E
$$1$$
Solution
- Probability that the sum of two dice is greater than $$3$$ is $$1$$ minus probability that the sum is less than or equal to $$3$$.
- probability that the sum is less than or equal to $$3$$ is $$\dfrac{3}{36} =\dfrac{1}{12}$$.
- So the probability that the sum is greater than $$3$$ is $$1-1/12 = 11/12$$.
Question 3
1 / -0

If A & B are two given events, then $$P(A \cap B)$$ is

A
Not less than $$P(A) + P(B) -1$$

B
Not greater than $$P(A) + P(B)-P(A\cup B)$$

C
Equal to $$P(A) + P(B) + P(A\cap B)$$

D
Equal to $$P(A) + P(B) + P(A\cup B)$$

Solution

We have $$P(A\cup B) \le 1$$
$$P(A)+P(B) - P(A\cap B) ≤ 1$$
$$\Rightarrow P(A\cap B) ≥ P(A) + P(B) - 1$$
$$P(A\cap B)$$ is not less than $$P(A)+P(B) -1$$
Question 4
1 / -0

If a leap year is selected at random what is the probability that it will contain $$53$$ Tuesdays?

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

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

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

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

Solution

$$1 year = 365$$ days
A leap year has $$366$$ days
A year has $$52$$ weeks. Hence there will be 52 Tuesdays for sure.
$$52 weeks = 52 \times 7 = 364 days$$
$$366 – 364 =2$$ days
In a leap year there will be 52 Tuesdays and 2 days will be left.
These 2 days can be:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Of these total 7 outcomes, the favorable outcomes are 2.
Hence the probability of getting 53 Tuesdays in a leap year $$P(E)=\frac{2}{7}$$
Question 5
1 / -0

In a group of $$10$$ people, $$70$$% take vitamins. If you randomly choose $$2$$ of them, what is the probability that neither person selected takes vitamins?

A
$$.0677$$

B
$$.09$$

C
$$.30$$

D
$$.47$$

E
$$.49$$

Solution

Total no. of people $$=10$$
$$\Rightarrow 70$$% of people take vitamins, i.e, $$\dfrac{70}{100}\times10=7$$ people.
$$\Rightarrow 30$$% of people doesn't take vitamins, i.e, $$\dfrac{30}{100}\times10=3$$ people.

$$\Rightarrow P(2$$ person who don't take vitamins $$) =\dfrac{^3C_2}{^{10}C_2}=0.0677.$$
Hence, the answer is $$0.0677.$$
Question 6
1 / -0

A game has 2 spinners. Spinner #1 has a probability of landing red of $$2/3$$. And, spinner#2 has a probability of landing red of $$1/5$$.
What is the probability spinner#1 lands red AND spinner#2 does NOT land red?

A
$$2/15$$

B
$$8/15$$

C
$$13/15$$

D
$$1/5$$

E
$$3/5$$
Solution
- probability of spinner#1 lands red AND spinner#2 doesnot land red is $$2/3 \times (1-1/5) = 2/3 \times 4/5 = 8/15$$
Question 7
1 / -0

If $$P(A)= \dfrac{1}{3} , P (B) = \dfrac{1}{2}$$ and A, B are mutually exclusive, find $$P(A' \cap B')$$.

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

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

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

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

Solution

Given:
$$P(A)=\dfrac 13, P(B)=\dfrac 12$$
A and B are mutually exclusive, hence
$$P(A\cup B)=P(A)+P(B)=\dfrac 13+\dfrac 12=\dfrac 56$$
And,
$$P(A'\cap B')=1−P(A\cup B)=1-\dfrac 56=\dfrac 16$$
Question 8
1 / -0

If $$P(A)=\frac{1}{8}$$ and $$P(B)=\frac{5}{8}$$. Which of the following statements is correct?

A
$$P(A\cup B)\leq \frac{3}{4}$$

B
$$P(A\cap B)\leq \frac{5}{8}$$

C
$$P(\overline{A}\cap B)\leq \frac{5}{8}$$

D
$$P(A\cap B)\geq \frac{5}{8}$$

Solution

Given $$P(A)=\dfrac{1}{8}, P(B)=\dfrac{5}{8}$$.

We know that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$\Rightarrow P(A\cup B)=\dfrac{1}{8}+\dfrac{5}{8}-P(A\cap B)$$

$$=\dfrac{6}{8}-P(A\cap B)$$

$$=\dfrac{3}{4}-P(A\cap B)$$

Since $$P(A\cup B)$$ is $$\dfrac{3}{4}-$$ some positive quantity.

We can say that $$P(A\cup B) \leq \dfrac{3}{4}$$

Hence statement $$(A)$$ is correct.
Question 9
1 / -0

Let A and B be two events such that $$P(A\cup B) = P(A\cap B)$$. Then,
Statement 1: $$P(A\cap \overline{B})=P(\overline{A}\cap B)=0$$
Statement 2: If $$P(A)+P(B)=1$$.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

Solution

We have,
$$P(A\cup B) = P(A\cap B)$$
$$\Rightarrow P(A)+P(B)-P(A\cap B)=P(A\cap B)$$ ...(i)
$$\Rightarrow \{P(A)- P(A\cap B)\}+\{P(B)-P(A\cap B)\}=0$$
$$\Rightarrow P(A\cap \overline{B})+P(\overline{A}\cap B)=0$$
$$\Rightarrow P(A\cap \overline{B})=P(\overline{A}\cap B)=0$$
So, statement 1 is true.
From (i), we have
$$P(A)+P(B)=2P(A\cap B)$$
So, $$P(A)+P(B)=1$$ may not be true.
Question 10
1 / -0

If A and B are two events. The probability that at most one of A, B occurs is

A
$$1-P(A\cap B)$$

B
$$P(\overline{A})+P(\overline{B})-P(\overline{A}\cap {B})$$

C
$$P(\overline{A})+P(\overline{B})+P(\overline{A}\cap \overline{B})$$

D
All of the above

Solution

C is the event that exactly one of the following events:
A but not B
B but not A
Not B and Not A (Neither A nor B)
1 is the event $$A∖B$$ or $$A∩B^C$$
2 is the event $$B∖A$$or $$B∩A^C$$
3 is the event $$B^C∩A^C$$ (not B and not A) or $$(A∪B)^C (not (A or B))$$

Observe that the events are mutually exclusive.
Hence,
$$P(C)=P(A∩B^C)+P(B∩A^C)+P(B^C∩A^C)$$
$$P(C)=P(A∩B^C)+P(B∩A^C)+P(B^C∩A^C)$$
$$P(C)=1−P(A∩B)$$$
$$P(C)=1−P(A)P(B)$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 39

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Probability Test - 39

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SHARING IS CARING

Question 1

$$\dfrac{9}{100}$$

$$\dfrac{1}{100}$$

$$\dfrac{9}{10}$$

$$\dfrac{1}{10}$$

Solution

Question 2

$$1/4$$

$$3/4$$

$$5/6$$

$$11/12$$

$$1$$

Solution

Question 3

Not less than $$P(A) + P(B) -1$$

Not greater than $$P(A) + P(B)-P(A\cup B)$$

Equal to $$P(A) + P(B) + P(A\cap B)$$

Equal to $$P(A) + P(B) + P(A\cup B)$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$.0677$$

$$.09$$

$$.30$$

$$.47$$

$$.49$$

Solution

Question 6

$$2/15$$

$$8/15$$

$$13/15$$

$$1/5$$

$$3/5$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$$P(A\cup B)\leq \frac{3}{4}$$

$$P(A\cap B)\leq \frac{5}{8}$$

$$P(\overline{A}\cap B)\leq \frac{5}{8}$$

$$P(A\cap B)\geq \frac{5}{8}$$

Solution

Question 9

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Solution

Question 10

$$1-P(A\cap B)$$

$$P(\overline{A})+P(\overline{B})-P(\overline{A}\cap {B})$$

$$P(\overline{A})+P(\overline{B})+P(\overline{A}\cap \overline{B})$$

All of the above

Solution

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