Probability Test

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Probability Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

A box contains 3 white and 2 black balls. Two balls are drawn at random one after the other. If the balls are not replaced. what is the probability that both the balls are black ?

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

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

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

D
None of the above

Solution

Probabilty of drawing black in the first draw$$ =\cfrac {\text{total black balls} }{ \text{total balls} } =\cfrac { 2 }{ 5 }$$
Probability of drawing black in the second draw $$ =\cfrac { 1 }{ 4 } $$
Probability that both balls are black $$=$$ Probability of first ball being black $$\times $$ Probability of second ball being black
$$=\cfrac { 2 }{ 5 } \times \cfrac { 1 }{ 4 } \\ =\cfrac { 1 }{ 10 } $$
Question 2
1 / -0

Two events A and B will be independent if

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

B
$$P(A) + P(B) = 1$$

C
$$P(A) = P(B)$$

D
$$A$$ and $$B$$ are mutually exclusive

Solution

Two events A & B will be independent if
$$P\left( A\cap { B } \right) =P\left( A \right) \cap { P\left( B \right) } $$
If A & B are independent, then $${ A }^{ ' }\& \ { B }^{ ' }$$ are also independent
$$\therefore P\left( { A }^{ ' }\cap { { B }^{ ' } } \right) =P\left( { A }^{ ' } \right) P\left( { B }^{ ' } \right) $$
$$=\left( 1-P(A) \right) \left( 1-P(B) \right) $$
Question 3
1 / -0

The probability distribution of a discrete random variable $$X$$ is:
$$X = x$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
$$P(X = x)$$ $$k$$ $$2k$$ $$3k$$ $$4k$$ $$5k$$
Find $$P (X\leq 4)$$

A
$$\cfrac 23$$

B
$$\cfrac 34$$

C
$$\cfrac 45$$

D
$$\cfrac 56$$

Solution

Here, we must have $$k + 2k + 3k + 4k + 5k = 1$$
$$\Rightarrow 15k = 1$$
$$\Rightarrow k = \dfrac {1}{15}$$
Now, $$P(X \leq 4) = k + 2k + 3k + 4k$$
$$= \dfrac {1}{15} + \dfrac {2}{15} + \dfrac {3}{15} + \dfrac {4}{15}$$
$$= \dfrac {10}{15}$$
$$\therefore P(X\leq 4) = \dfrac {2}{3}$$
Question 4
1 / -0

An urn contains $$5$$ red and $$2$$ black balls. Two balls are randomly drawn. Let $$X$$ represent the number of black balls. What are the possible values of $$X$$ ? Is $$X$$ a random variable ?

A
$$0,1,2$$

B
$$3,5,7$$

C
$$7,7,8$$

D
$$1,5,7$$

Solution

The two balls selected can be represented as $$BB, BR, RB, RR$$, where $$B$$ represents a black ball and $$R$$ represents a red ball.
$$X$$ represents the number of black balls.
$$\therefore X \left(BB\right) = 2$$
$$X \left(BR\right) = 1$$
$$X \left(RB\right) = 1$$
$$X \left(RR\right) = 0$$
Therefore, the possible values of $$X$$ are $$0, 1$$ and $$2$$.
Yes, $$X$$ is a random variable.
Question 5
1 / -0

If $$P(A/ B) = P(A)$$, then

A
$$A$$ is independent of $$B$$

B
$$B$$ is independent of $$A$$

C
$$B$$ is dependent of $$A$$

D
Both (a) and (b)

Solution

If P(A|B) = P(A), then events A and B are said to be independent, i.e., A is independent on B and vice versa.
Question 6
1 / -0

The probability that a leap year will have $$53$$ sundays is

A
$$1/7$$

B
$$2/7$$

C
$$5/7$$

D
$$6/7$$

Solution

A leap year has $$52$$ weeks and $$2$$ days.
The $$53^{rd}$$ Sunday will be from these extra two days.
These $$2$$ days can be (Sunday,Monday) or (Mon,Tue) or (Tue,Wed).....(Sat,Sun)
There are $$7$$ possibilities for these $$2$$ days
Out of which Sunday is coming in $$2$$ possibilities
$$\therefore P(\text{2 sundays in leap year})=\dfrac{2}{7}$$
Question 7
1 / -0

The probability of A's a Tenis game against $$B$$ is $$\dfrac{1}{3}$$. Find the probability that A wins at least $$1$$ of the total $$3$$ set of games.

A
$$\dfrac{1}{19}$$

B
$$\dfrac{19}{27}$$

C
$$\dfrac{1q}{19}$$

D
$$\dfrac{7}{19}$$

Solution

Given: Probability of $$A$$ winning the tennis game against $$B$$ is $$\dfrac 13$$ and a total of $$3$$ games is played.
To find: the probability that $$A$$ wins at least $$1$$ of the total $$3$$ set of games
Probability of $$A$$ winning at least $$1$$ game $$=$$ P (A winning exactly $$1$$ game) $$+$$ P(A winning exactly $$2$$ games) $$+$$ P(A winning exactly $$3$$ games)
This is Bernoulli's' theorem, hence
P(Success in each trial) $$={}^nC_k p^k(1-p)^{n-k}$$, in this case $$n = 3,$$ $$p=\dfrac 13$$ and $$k = 1, 2, 3$$
$$= {}^3C_1\left( \dfrac 13\right)^1\left(1-\dfrac 13\right)^{3-1}+{}^3C_2\left( \dfrac 13\right)^2\left(1-\dfrac 13\right)^{3-2}+{}^3C_3\left( \dfrac 13\right)^3\left(1-\dfrac 13\right)^{3-3}$$

$$= 3\times \dfrac 13\times \dfrac {2^2}{3^2}+3\times \dfrac {2}{3^2}\times \dfrac 1{3}+1\times \dfrac 1{3^3}\times 1$$

$$= \dfrac{12}{27}+\dfrac{6}{27}+\dfrac{1}{27}=\dfrac {12+6+1}{27}=\dfrac{19}{27}$$
Question 8
1 / -0

One percent of the population suffers from a certain disease. There is blood test for this disease, and it is $$99\%$$ accurate, in other words, the probability that is gives the correct answer is $$0.99$$, regardless of whether the person is sick or healthy. A person takes the blood test, and the result says that he has the disease. The probability that he actually has the disease, is?

A
$$0.99\%$$

B
$$25\%$$

C
$$50\%$$

D
$$75\%$$

Solution

A be the event of having the disease.
B be the event of testing positive.
$$P(A) = 0.01$$
$$P(B) = P(B/A)P(A) + P(B/ not A)P(not A)$$
$$P(B) = 0.01*0.99+0.01*0.99 = 0.0198$$
$$P(A \cap B) = 0.99*0.01$$
$$P(A/B) = \cfrac{P(A\cap B)}{P(B)}$$
$$\implies \cfrac{0.99*0.01}{0.0198}$$
=$$0.5$$ or $$50\%$$
Question 9
1 / -0

Box I contains $$2$$ white and $$3$$ red balls and box II contains $$4$$ white and $$5$$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?

A
$$\cfrac{54}{44}$$

B
$$\cfrac{54}{14}$$

C
$$\cfrac{54}{104}$$

D
None of these

Solution

Probability that the ball drawn is red and from ! =$$P(R/A)$$
$$P(R/A) = \cfrac{P(A/R)\times P(A)}{P(B/R)\times P(B) + P(A/R) \times P(A)}$$
$$P(R/A) = \cfrac{3}{5}, P(R/B) = \cfrac{5}{9}$$
$$P(A) = \cfrac{1}{2}, P(B) = \cfrac{1}{2}$$
$$P(R/A) = \cfrac{\cfrac{3}{5}\times \cfrac{1}{2}}{\cfrac{5}{9}\times \cfrac{1}{2} + \cfrac{3}{5} \times \cfrac{1}{2}} = \cfrac{54}{104}$$
Question 10
1 / -0

In a simultaneous throw of a pair of dice, find the probability of getting:
A doublet odd numbers

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

B
$$\dfrac {1}{36}$$

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

D
$$\dfrac {5}{12}$$

Solution

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Re-Attempt Weekly Quiz Competition

$$X = x$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$
$$P(X = x)$$	$$k$$	$$2k$$	$$3k$$	$$4k$$	$$5k$$

Probability Test - 16

Result

Probability Test - 16

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SHARING IS CARING

Question 1

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

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

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

None of the above

Solution

Question 2

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

$$P(A) + P(B) = 1$$

$$P(A) = P(B)$$

$$A$$ and $$B$$ are mutually exclusive

Solution

Question 3

$$\cfrac 23$$

$$\cfrac 34$$

$$\cfrac 45$$

$$\cfrac 56$$

Solution

Question 4

$$0,1,2$$

$$3,5,7$$

$$7,7,8$$

$$1,5,7$$

Solution

Question 5

$$A$$ is independent of $$B$$

$$B$$ is independent of $$A$$

$$B$$ is dependent of $$A$$

Both (a) and (b)

Solution

Question 6

$$1/7$$

$$2/7$$

$$5/7$$

$$6/7$$

Solution

Question 7

$$\dfrac{1}{19}$$

$$\dfrac{19}{27}$$

$$\dfrac{1q}{19}$$

$$\dfrac{7}{19}$$

Solution

Question 8

$$0.99\%$$

$$25\%$$

$$50\%$$

$$75\%$$

Solution

Question 9

$$\cfrac{54}{44}$$

$$\cfrac{54}{14}$$

$$\cfrac{54}{104}$$

None of these

Solution

Question 10

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

$$\dfrac {1}{36}$$

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

$$\dfrac {5}{12}$$

Solution

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