Probability Test

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Probability Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Suppose that two cards are drawn at random from a deck of cards. Let $$X$$ be the number of aces obtained. Then the value of $$E\left(X\right)$$ is

A
$$\displaystyle\frac{37}{221}$$

B
$$\displaystyle\frac{5}{13}$$

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

D
$$\displaystyle\frac{2}{13}$$

Solution

Let $$X$$ denote the number of aces obtained. Therefore, $$X$$ can take any of the values of 0, 1, or 2.
In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.
$$\therefore P\left( X=0 \right) =P\left( 0 ace and 2 non-ace cards \right) =\displaystyle\frac { _{ }^{ 4 }{ { C }_{ 0 } }\times _{ }^{ 48 }{ { C }_{ 2 } } }{ _{ }^{ 52 }{ { C }_{ 2 } } } =\displaystyle\frac { 1128 }{ 1326 } $$
$$P\left( X=1 \right) =P\left( 1 ace and 1 non-ace cards \right) =\displaystyle\frac { _{ }^{ 4 }{ { C }_{ 1 } }\times _{ }^{ 48 }{ { C }_{ 1 } } }{ _{ }^{ 52 }{ { C }_{ 2 } } } =\displaystyle\frac { 192 }{ 1326 } $$
$$P\left( X=2 \right) =P\left( 2 ace and 0 non-ace cards \right) =\displaystyle\frac { _{ }^{ 4 }{ { C }_{ 2 } }\times _{ }^{ 48 }{ { C }_{ 0 } } }{ _{ }^{ 52 }{ { C }_{ 2 } } } =\displaystyle\frac { 6 }{ 1326 } $$
Thus, the probability distribution is as follows.
Then, $$E\left( X \right) =\sum { { p }_{ i }{ x }_{ i } } $$
$$=0\times \displaystyle\frac { 1128 }{ 1326 } +1\times \displaystyle\frac { 192 }{ 1326 } +2\times \displaystyle\frac { 6 }{ 1326 } $$
$$=\displaystyle\frac { 204 }{ 1326 } $$
$$=\displaystyle\frac { 2 }{ 13 } $$
Therefore, the correct answer is D.
Question 2
1 / -0

A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one ball is drawn from each bag, find the probability that both are green.

A
13/70

B
1/4

C
6/35

D
8/35

Solution

$$Bag\quad A\quad has\quad 4\quad green\quad balls\quad and\quad 6\quad red\quad balls\quad \\ \Rightarrow probability\quad of\quad choosing\quad green\quad ball\quad from\quad A\quad is\quad p({ green }_{ A })=\frac { 4 }{ 10 } \\ Bag\quad A\quad has\quad 3\quad green\quad balls\quad and\quad 4\quad red\quad balls\quad \\ \Rightarrow probability\quad of\quad choosing\quad green\quad ball\quad from\quad B\quad is\quad p({ green }_{ B })=\frac { 3 }{ 7 } \\ On\quad choosing\quad one\quad ball\quad from\quad each\quad bag\quad probability\quad that\quad both\quad are\quad green=\quad p({ green }_{ A })*p({ green }_{ B })\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\frac { 4 }{ 10 } *\frac { 3 }{ 7 } =\frac { 6 }{ 35 } $$
Question 3
1 / -0

A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are 100 Year-III, 150 Year-II and 200 Year-I students who applied. Each Year-III's name is placed in the lottery 3 times; each Year-II's name, 2 times and Year-I's name, 1 time. What is the probability that a Year-III's name will be chosen?

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

B
$$\frac{2}{8}$$

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

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

Solution

Total names in the lottery
$$=3\times 100+2\times 150+200=800$$
Number of Year-III's names$$=3\times 100=300$$
Required probability $$=\frac{300}{800}=\frac{3}{8}$$
Question 4
1 / -0

In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

A
0.43

B
0.40

C
0.67

D
0.60

Solution

$$P(M\, and\, S) = 0.40$$
$$P(M) = 0.60$$
$$P(S|M) = \dfrac{P(M \,and\, S)}{P(S)} =\dfrac{0.40}{0.60} =\dfrac{2}{3} = 0.67$$
Question 5
1 / -0

If $$P\left( A|B \right) > P\left( A \right) $$, then which of the following is correct

A
$$P\left( B|A \right) < P\left( B \right) $$

B
$$P\left( A\cap B \right) < P\left( A \right) \cdot P\left( B \right) $$

C
$$P\left( B|A \right) > P\left( B \right) $$

D
$$P\left( B|A \right) =P\left( B \right) $$

Solution

$$P\left( A|B \right) >P\left( A \right) $$
$$\Rightarrow \displaystyle\frac { P\left( A\cap B \right) }{ P\left( B \right) } >P\left( A \right) $$
$$\Rightarrow P\left( A\cap B \right) >P\left( A \right) \cdot P\left( B \right) $$
$$\Rightarrow \displaystyle\frac { P\left( A\cap B \right) }{ P\left( A \right) } >P\left( B \right) $$
$$\Rightarrow P\left( B|A \right) >P\left( B \right) $$
Thus, the correct answer is C.
Question 6
1 / -0

A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag . Find the probability that one ball is red and one is green ?

A
19/20

B
17/20

C
8/10

D
21/40

E
15/40

Solution

We have,
Bag 1:
Red ball = 5
Green ball = 3
Bag 2:
Red ball = 4
Green ball = 6
Now, according to the question we will have to cases.
1. Probability of getting Red ball from bag 1 and green ball from bag 2 = (5/8) * (6/10) = 30/80
2. Probability of getting Green from bag 1 and Red from bag 2 = (3/8) * (4/10) = 12/80
Therefore, probability of getting Red and Green,
= (30/80) + (12/80) = 42/80 = 21/40.
Question 7
1 / -0

A speaks truth in 75% of cases and B in 80% of cases. In what percent of cases are they likely to contradict each other in narrating the same event?

A
$$20\%$$

B
$$25\%$$

C
$$30\%$$

D
$$35\%$$

Solution

Different possible cases of contradiction,
A speaks truth and B does not speaks truth.
Or, A does not speak truth and B speaks truth.
$$\displaystyle =\left(\frac{3}{4}\times \frac{1}{5}\right)+\left(\frac{1}{4}\times \frac{4}{5}\right)=\frac{3}{20} + \frac{4}{20}=\frac{7}{20} =35\%$$
Question 8
1 / -0

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are $$\displaystyle \frac { 1 }{ 4 } ,\frac { 1 }{ 2 } $$ and $$\displaystyle \frac { 5 }{ 8 } $$, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A
$$\displaystyle \frac { 11 }{ 8 } $$

B
$$\displaystyle \frac { 7 }{ 8 } $$

C
$$\displaystyle \frac { 9 }{ 64 } $$

D
$$\displaystyle \frac { 5 }{ 64 } $$

E
$$\displaystyle \frac { 3 }{ 64 } $$

Solution

P(Xavier will solve)$$=\dfrac{1}{4}$$
P(Yvonne will solve)$$=\dfrac{1}{2}$$
P(Zelda will NOT solve) $$= 1- \dfrac{5}{8} = \dfrac{3}{8}.$$

Now, we need to multiply all this Ps to find an answer:
$$p= (\dfrac{1}{4})*(\dfrac{1}{2})*(\dfrac{3}{8}) = \dfrac{3}{64}.$$
Question 9
1 / -0

A parents has two children. If one of them is boy, then the probability that other is, also a boy, is

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

B
$$\dfrac{1}{4}$$

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

D
None of these

Solution

Total cases are $$BB, BG, GB, GG = 4$$
Favorable cases are $$BB, BG, GB = 3$$
Let $$P\left( A \right) =$$ Probability of a boy in two children
$$=\dfrac { 3 }{ 4 } $$
The probability that the second child is also a boy is $$P\left( A\cap B \right) =\dfrac { 1 }{ 4 } $$
We have to find $$P\left( \dfrac{ B }{ A } \right) =\dfrac { P\left( A\cap B \right) }{ P\left( A \right) } =\dfrac { \dfrac { 1 }{ 4 } }{ \dfrac{ 3 }{ 4 } } =\dfrac { 1 }{ 3 } $$
Question 10
1 / -0

If $$A$$ and $$B$$ are two events such that $$P\left( A \right) \neq 0$$ and $$P\left( B|A \right) =1$$, then

A
$$A\subset B$$

B
$$B\subset A$$

C
$$B=\phi $$

D
$$A=\phi $$

Solution

$$P\left( A \right) \neq 0$$ and $$P\left( B|A \right) =1$$
$$P\left( B|A \right) =\displaystyle\frac { P\left( B\cap A \right) }{ P\left( A \right) } $$
$$1=\displaystyle\frac { P\left( B\cap A \right) }{ P\left( A \right) } $$
$$P\left( A \right) =P\left( B\cap A \right) $$
$$\Rightarrow A\subset B$$
Thus, the correct answer is A.

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Re-Attempt Weekly Quiz Competition

Probability Test - 26

Result

Probability Test - 26

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Question 1

$$\displaystyle\frac{37}{221}$$

$$\displaystyle\frac{5}{13}$$

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

$$\displaystyle\frac{2}{13}$$

Solution

Question 2

13/70

1/4

6/35

8/35

Solution

Question 3

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

$$\frac{2}{8}$$

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

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

Solution

Question 4

0.43

0.40

0.67

0.60

Solution

Question 5

$$P\left( B|A \right) < P\left( B \right) $$

$$P\left( A\cap B \right) < P\left( A \right) \cdot P\left( B \right) $$

$$P\left( B|A \right) > P\left( B \right) $$

$$P\left( B|A \right) =P\left( B \right) $$

Solution

Question 6

19/20

17/20

8/10

21/40

15/40

Solution

Question 7

$$20\%$$

$$25\%$$

$$30\%$$

$$35\%$$

Solution

Question 8

$$\displaystyle \frac { 11 }{ 8 } $$

$$\displaystyle \frac { 7 }{ 8 } $$

$$\displaystyle \frac { 9 }{ 64 } $$

$$\displaystyle \frac { 5 }{ 64 } $$

$$\displaystyle \frac { 3 }{ 64 } $$

Solution

Question 9

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

$$\dfrac{1}{4}$$

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

None of these

Solution

Question 10

$$A\subset B$$

$$B\subset A$$

$$B=\phi $$

$$A=\phi $$

Solution

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