Probability Test 17

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Probability Test 17

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Weekly Quiz Competition

Question 1
1 / -0

The different possibilities of distribution of $$2$$ black and $$2$$ white balls in two containers are as shown. Which of the following statements is true?

A
(iv) has the maximum possibility of picking a black ball

B
(i) and (iv) has equal probability of picking a white ball

C
(iii) has the maximum possibility of picking a white ball

D
(ii) has the maximum possibility of picking a white ball

Solution

(i) $$P(B)= \dfrac{1}{2}(\dfrac{1}{2})+\dfrac{1}{2}(\dfrac{1}{2})=\dfrac{1}{2}$$
(ii) $$P(B)=\dfrac{1}{2}$$
(iii $$P(B))= \dfrac{1}{2}(\dfrac{2}{3})=\dfrac{1}{3}$$
(iv) $$P(B)= \dfrac{1}{2}+\dfrac{1}{2}(\dfrac{1}{3})=\dfrac{2}{3}$$
Question 2
1 / -0

Let A and B be two events such that $$P(\overline{A\cup B})=\dfrac{1}{6}, P(A\cap B)=\dfrac{1}{4}$$ and $$P(\overline{A})=\dfrac{1}{4}$$, where $$\overline{A}$$ stands for the complement of the event A. Then the events A and B are?

A
Independent but not equally likely

B
Independent and equally likely

C
Mutually exclusive and independent

D
Equally likely but not independent

Solution

P(A')=1/4 so P(A)=3/4
P(AUB)'=1/6, P(AUB)=5/6
P(A∩B)=1/4
P(AUB)=P(A)+P(B)-P(A∩B)=3/4+P(B)-1/4
5/6=1/2+P(B)
P(B)=2/6=1/3
P(A)*P(B)=1/3*3/4=1/4=P(A∩B)
A and B are independent but not equally likely
Question 3
1 / -0

If A and B are two mutually exclusive events, then?

A
$$P(A) \leq P(B')$$

B
$$P(A)\leq P(B)$$

C
$$P(A)\geq P(B')$$

D
None of these

Solution

$$\begin{matrix} P(A\cap B)=0 \\ P(A\cap B)=P(A)+P(B)-P(A\cup B) \\ 0=P(A)+P(B)-P(A\cup B) \\ P(A)+P(B)=P(A\cup B) \\ P(A)=P(A\cup B)-P(B) \\ P(A)\leqslant P(B) \\ P(A)\leqslant P(B)\, \, \, \, \, \, Ans.\, \, \, \, \, \, \, \, \, \\ \therefore obabilty\, \, of\, \, any\, \, even\, \, is\, \, 0\, or\, \, greater\, \, than\, \, zero. \\ \\ \end{matrix}$$
Question 4
1 / -0

A special die with number $$1$$, $$-1$$, $$2$$, $$-2$$, $$0$$ and $$3$$ is thrown thrice. The probability that total is $$6$$ is

A
$$\dfrac{1}{{108}}$$

B
$$\dfrac{{25}}{{216}}$$

C
$$\dfrac{5}{{216}}$$

D
$$\dfrac{5}{{108}}$$

Solution

outcomes are 1,-1.2,-2,0,3

For Sum to be 6 -

2,2,2 1 case

3,3,0 3!/2= 3 cases

1,2,3 3!=6 cases
Total favourable outcomes=1+3+6=10
Total outcomes =6$$\times$$6$$\times$$6=216
Probability=10/216=5/108
Question 5
1 / -0

The number of ways in which $$6$$ men can be arranged in a row, so that three particular men are consecutive, is

A
$$4! \times 3!$$

B
$$4!$$

C
$$3! \times 3!$$

D
none of these

Solution

To arrange $$6$$ men in a row such three particular men are consecutive $$M_1M_2M_3M_4M_5M_6$$
three men to be consecutive lets make $$3$$ men in a Group
$$(M_1M_2M_3)$$ $$M_4M_5M_6$$
Total no. of ways of arranging them is $$4!\times3!$$.
Question 6
1 / -0

If $$\displaystyle \frac{1+3p}{3}, \frac{1-p}{4}$$ and $$\dfrac{1-2p}{2}$$ are the probabilities of the three mutually exclusive events, then $$p \in $$

A
$$[0, 1]$$

B
$$\left [0, \dfrac{1}{2}\right]$$

C
$$\left [\dfrac{1}{3}, 1\right]$$

D
$$\left[\dfrac{1}{3}, \dfrac{1}{2}\right]$$

Solution
Question 7
1 / -0

If $$\frac { 1+4p }{ 4 } ,\frac { 1-p }{ 4 } ,\frac { 1-2p }{ 4 } $$ are probabilities of three mutually exclusive and exclusive and exhaustive events, then the possible value of p belong to the set

A
$$\left( 0,\frac { 2 }{ 3 } \right) $$

B
$$\left[ 0,\frac { 1 }{ 2 } \right] $$

C
$$\left[ -\frac { 1 }{ 4 } ,\frac { 1 }{ 2 } \right] $$

D
$$\left[ -\frac { 2 }{ 3 } ,\frac { 2 }{ 3 } \right] $$

Solution

Since the probabilities are greater than 0 or less than (=)1.
Also, the probabilities are mutually excessive
and exhaustive.
$$ 0< \dfrac{1+4p}{4}< 1 $$ or $$0< 1+4p< 4 ...(1)$$
$$ 0< \dfrac{1-p}{4} < 1$$
$$ 0< 1-p< 4 ...(2)$$
$$0< 1 \dfrac{1-2p}{4}< 1$$
$$ 0< 1 -2p< 4 ...(3)$$
from (1), (2), (3), we get
$$ -\dfrac{1}{4} < p < \dfrac{3}{4}$$
$$ 1> p> -3 $$ or $$ -3 < p< 1 $$
$$ -\dfrac{3}{2} < p< \dfrac{1}{2}$$
$$ \therefore $$ interval of p can be $$ [\dfrac{-1}{4}, \dfrac{1}{2}]$$
Option C
Question 8
1 / -0

In a class 5% of boys and 10% of girls have an I.Q of more than 150.In this class 60% of students are boys. If a student is selected at random and found to have an I.Q. of more than 150. Find the probability that the student is a boy.

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

B
$$\dfrac{23}{7}$$

C
$$\dfrac{3}{5}$$

D
None of these

Solution

Let us consider the problem
$$E_1$$ : Event that boys are selected
$$E_2$$ : Event that girls are selected
$$A$$ : event that have IQ $$150$$
Implies that,
\begin{array}{l} P\left( { A/{ E_{ 1 } } } \right) =\dfrac { 5 }{ { 100 } } \\ P\left( { A/{ E_{ 2 } } } \right) =\dfrac { { 10 } }{ { 100 } } \\ P\left( { { E_{ 1 } } } \right) =\dfrac { { 60 } }{ { 100 } } ,P\left( { { E_{ 2 } } } \right) =\dfrac { { 40 } }{ { 100 } } \\ P\left( { A|{ E_{ 1 } } } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right) } }{ { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { A|{ E_{ 2 } } } \right) } } \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } } }{ { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } +\dfrac { { 40 } }{ { 100 } } \times \dfrac { { 10 } }{ { 100 } } } } \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 60\times 5 } }{ { 60\times 5+40\times 10 } } \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 300 } }{ { 300+400 } } \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { 3 }{ 7 } \end{array}

Hence, the probability is $$\dfrac {3}{7}$$
Question 9
1 / -0

Assume that the birth of a boy or girl to a couple to be equally likely ,mutually exclusive,independent of the other children in the family .for a couple having $$6$$ children ,the probability that their "three oldest are boys" is

A
$$\dfrac{20}{{64}}$$

B
$$\dfrac{1}{{64}}$$

C
$$\dfrac{2}{{64}}$$

D
$$\dfrac{8}{{64}}$$

Solution

$$ probability\,of\,son=\frac{1}{2} $$
$$ probability\,of\,girl=\frac{1}{2} $$
$$ from\text{ 6 children probability of 3 oldest are boys is} $$
$$ \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8} $$
Question 10
1 / -0

There are $$n$$ distinct white and $$n$$ distinct black balls. The number of ways of arranging them in a row so that neighbouring balls are of different colours is:

A
$$n+1C_{2}$$

B
$$(2)(n\ !)^{2}$$

C
$$\dfrac{(n\ !)}{2}$$

D
$$none\ of\ these$$

Solution

Possible arrangements are BWBW....... or WBWB......
For combination BWBW..... , n blacks can permutate in n! ways and n white balls can permutate in n! ways
Total number of arrangements are (n!)(n!)
Since there are two possible arrangements total ways =$$2{ (n!) }^{ 2 }$$

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Probability Test 17

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Probability Test 17

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Question 1

(iv) has the maximum possibility of picking a black ball

(i) and (iv) has equal probability of picking a white ball

(iii) has the maximum possibility of picking a white ball

(ii) has the maximum possibility of picking a white ball

Solution

Question 2

Independent but not equally likely

Independent and equally likely

Mutually exclusive and independent

Equally likely but not independent

Solution

Question 3

$$P(A) \leq P(B')$$

$$P(A)\leq P(B)$$

$$P(A)\geq P(B')$$

None of these

Solution

Question 4

$$\dfrac{1}{{108}}$$

$$\dfrac{{25}}{{216}}$$

$$\dfrac{5}{{216}}$$

$$\dfrac{5}{{108}}$$

Solution

Question 5

$$4! \times 3!$$

$$4!$$

$$3! \times 3!$$

none of these

Solution

Question 6

$$[0, 1]$$

$$\left [0, \dfrac{1}{2}\right]$$

$$\left [\dfrac{1}{3}, 1\right]$$

$$\left[\dfrac{1}{3}, \dfrac{1}{2}\right]$$

Solution

Question 7

$$\left( 0,\frac { 2 }{ 3 } \right) $$

$$\left[ 0,\frac { 1 }{ 2 } \right] $$

$$\left[ -\frac { 1 }{ 4 } ,\frac { 1 }{ 2 } \right] $$

$$\left[ -\frac { 2 }{ 3 } ,\frac { 2 }{ 3 } \right] $$

Solution

Question 8

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

$$\dfrac{23}{7}$$

$$\dfrac{3}{5}$$

None of these

Solution

Question 9

$$\dfrac{20}{{64}}$$

$$\dfrac{1}{{64}}$$

$$\dfrac{2}{{64}}$$

$$\dfrac{8}{{64}}$$

Solution

Question 10

$$n+1C_{2}$$

$$(2)(n\ !)^{2}$$

$$\dfrac{(n\ !)}{2}$$

$$none\ of\ these$$

Solution

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