Probability Test

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Probability Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

Out of $$2$$ men and $$3$$ women a team of two persons is to be formed such that there is exactly one man and one woman. Write the sample space of this experiment, then the total no. of combinations of team possible are

A
$$9$$

B
$$6$$

C
$$4$$

D
$$12$$

Solution

Given,
A team contains 2men and 3 women.
But 2 persons have to be selected such that there is exactly 1 men and 1 women.
The no.of combinations$$=^{2}C_{1}*^{3}C_{1}=6$$ (i.e., selecting 1 men out of 2 men=$$^{2}C_{1}$$,selecting 1woment out of 3 women=$$^{3}C_{1}$$)
Sample space={$$M1W1,M1W2,M1W3,M2W1,M2W2,M2W3$$}
Question 2
1 / -0

One card is drawn from a pack of $$52$$ cards.The probability of getting a $$10$$ of black suit is

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

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

C
$$ \displaystyle \frac{3}{26} $$

D
None

Solution

Favourable number of outcomes, with $$10$$ of black suit $$= 2$$
Total number of outcomes $$= 52$$
Thus, probability $$=\dfrac{2}{52} =\dfrac{1}{26}$$
Question 3
1 / -0

Two unbiased coins are tossed simultaneously. The probability of getting two heads

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

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

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

D
None

Solution

Favourable number of outcomes, getting two heads = $$1$$ i.e {HH}
Total number of outcomes =$$6$$ i.e {HH, HT, TH, TT}
Probability $$=\dfrac{1}{4}$$
Question 4
1 / -0

$$ P\left ( E \right )+P\left ( \bar{E} \right ) $$ is equal to

A
$$0$$

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

C
$$1$$

D
None

Solution

$$ P\left ( E \right )+P\left ( \bar{E} \right ) $$ = $$ P\left ( E \right )+ 1- P\left ( E \right ) $$
= $$1$$
Question 5
1 / -0

If A and B are two mutually exclusive and exhaustive events with $$P(B)=3P(A),$$ then what is the value of $$P(\overline{B})$$?

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

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

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

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

Solution

Since P(A) and P(B) are two mutually exclusive and exhaustive events, $$P(A)+P(B)=1$$ and $$P(B)=3P(A)$$
$$\Rightarrow\:P(A)+3P(A)=1$$
$$\Rightarrow\:4P(A)=1$$
$$\Rightarrow\:P(A)=\dfrac{1}{4}$$
$$\Rightarrow\:P(B)=1-\dfrac{1}{4}=\dfrac{3}{4}$$
$$\Rightarrow\:$$ $$P(\overline{B})$$ $$=1-\dfrac{3}{4}=\dfrac{1}{4}$$.
Question 6
1 / -0

There are $$50$$ students in a class and their results is below:

Result (Pass/Fail)

Pass

Fail

No. of students

$$35$$

$$15$$

If a student chosen at random out of the class (i.e., without any bias), find the probability that the student is not failing (i.e., the student passed the examination).

A
$$0.7$$

B
$$0.66$$

C
$$0.9$$

D
$$1$$

Solution

Total number of students $$= 50$$
Number of students passed $$= 35$$
Probability, student is not failed $$=$$ probability that student is passed $$= \dfrac{35}{50} = 0.7$$
Question 7
1 / -0

Choose the correct alternative for each of the following. If $$ \displaystyle P\left ( E_{1} \right )=\frac{1}{6} $$, $$ \displaystyle P\left ( E_{2} \right )=\frac{1}{3} $$, $$ \displaystyle P\left ( E_{3} \right )=\frac{1}{6} $$, where $$ E_{1} $$, $$ E_{2} $$, $$ E_{3} $$, $$ E_{4} $$ are elementary events of a random experiment, then P($$ E_{4} $$) is equal to

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

B
$$ \displaystyle \frac{2}{3} $$

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

D
None

Solution

Given, probabilities of different elements of the event:
$$ \displaystyle P\left ( E_{1} \right )=\frac{1}{6} $$, $$ \displaystyle P\left ( E_{2} \right )=\frac{1}{3} $$, $$ \displaystyle P\left ( E_{3} \right )=\frac{1}{6} $$
The sum of individual probabilities $$= 1$$
$$P (E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) = 1$$
$$\displaystyle \frac{1}{6} + \frac{1}{3} + \frac{1}{6} + P(E_{4}) = 1$$
$$\displaystyle \frac{2}{3} + P(E_{4}) = 1$$
$$P(E_{4}) = \dfrac{1}{3}$$
Question 8
1 / -0

The probability of getting number 10 in a throw of a dice is ____.

A
0

B
1

C
0.5

D
0.75

Solution

Since the outcome of a throw of dice can never be 10, the probability is 0.
Question 9
1 / -0

In a simultaneous throw of two dice what is the probability of getting a doublet ?

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

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

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

D
$$\displaystyle \frac{2}{3}$$

Solution

Total number of possibilities = $$36$$
Number of doublet = $$6$$
Thus, probability = $$\dfrac{6}{36} = \dfrac{1}{6}$$
Question 10
1 / -0

Two unbiased coins are tossed simultaneously. Find the probability of getting at most one head.

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

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

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

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

Solution

Since, Total possibilities are $$=$$ $$\{HH,HT,TH,TT\}$$
no. of cases with atmost one head are $$= \{HT,TH,TT\}$$
$$\therefore $$ probability = $$\dfrac {3}{4} $$
Option C is correct.

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Result (Pass/Fail)	Pass	Fail
No. of students	$$35$$	$$15$$

Probability Test - 12

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Probability Test - 12

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Question 1

$$9$$

$$6$$

$$4$$

$$12$$

Solution

Question 2

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

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

$$ \displaystyle \frac{3}{26} $$

None

Solution

Question 3

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

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

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

None

Solution

Question 4

$$0$$

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

$$1$$

None

Solution

Question 5

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

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

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

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

Solution

Question 6

$$0.7$$

$$0.66$$

$$0.9$$

$$1$$

Solution

Question 7

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

$$ \displaystyle \frac{2}{3} $$

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

None

Solution

Question 8

0

1

0.5

0.75

Solution

Question 9

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

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

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

$$\displaystyle \frac{2}{3}$$

Solution

Question 10

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

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

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

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

Solution

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