Probability Test

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Probability Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If A and B are two independent events such that $$P\left( A \right) = \dfrac{1}{2}$$ and $$P\left( B \right) = \dfrac{1}{5}$$, then

A
$$P\left( {A \cup B} \right) = \dfrac{3}{5}$$

B
$$P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{2}$$

C
$$P\left( {\dfrac{A}{{A\,\,\, \cup \,\,B}}} \right) = \dfrac{5}{6}$$

D
$$P\left( {\dfrac{{A\,\,\, \cap \,\,B}}{{\overline A \,\,\, \cup \,\,\overline B }}} \right) = 0$$

Solution

Given A and B are independent $$\Rightarrow P\left (A\cap B\right) =P(A) P(B) $$

$$\Rightarrow P(A\cup B) =P(A)+P(B)-P(A\cap B) $$

$$\Rightarrow P(A\cup B) =P(A)+P(B)-P(A) P(B) $$

$$\Rightarrow P(A\cup B) =\dfrac {1}{2}+\dfrac {1}{5}-\dfrac {1}{10}$$

$$\Rightarrow P(A\cup B) =\dfrac {3}{5}$$
Question 2
1 / -0

If E & F are events with $$P(E) \leq P(F)$$ & $$P(E\cap F) > 0$$, then?

A
Occurrence of E $$\Rightarrow$$ occurrence of F

B
Occurrence of F $$\Rightarrow$$ occurrence of E

C
Non-occurrence of E $$\Rightarrow$$ non-occurrence of F

D
None of the above implications holds

Solution

P(E)≤P(F)
P(E∩F)>0
Since P(E∩F)>0, P(E) not equals to P(F) so occurrence of
E is not same as F
Also,1-P(E) not equals to 1-P(F) so non-occurrence of E is also
not equals to non- occurrence or occurrence of F
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

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 5
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 6
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 7
1 / -0

Four dice are rolled, then the probability that at least one digit on the dice must be repeated is

A
$$\dfrac{1}{18}$$

B
$$\dfrac{13}{18}$$

C
$$\dfrac{5}{18}$$

D
$$\dfrac{1}{9}$$

Solution

Counting cases where no digit is repeated choices for 1st dice $$=6$$
Choices for 2nd dice $$=5$$
Choices for 3trd dice $$=4$$
Choices for 4th dice $$=3$$
$$\Rightarrow$$ $$P=\cfrac{6\times 5\times 4\times 3}{6\times 6\times 6\times 6}$$
$$=\cfrac{5}{6}\times \cfrac{2}{3}\times \cfrac{1}{2}$$
$$=\cfrac{5}{1}$$
Required probability $$=1-P$$
$$=\cfrac{13}{18}$$
Question 8
1 / -0

The probability of complementary event of a certain event is

A
$$1$$

B
$$0$$

C
$$0.5$$

D
$$-1$$

Solution

Probability of certain event is 1 at its a certain event its possibility of happening is 100%
so probability of its complement is 0
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

If P(1)=0.4 , P(B)=0.6 and $$P(AB)=0.15$$, then the value of $$P(A|A'B')$$ is

A
$$\cfrac { 10 }{ 17 } $$

B
$$\cfrac { 1 }{ 17 } $$

C
$$\cfrac { 4 }{ 17 } $$

D
$$\cfrac { 5 }{ 17 } $$

Solution

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Probability Test - 49

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Probability Test - 49

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SHARING IS CARING

Question 1

$$P\left( {A \cup B} \right) = \dfrac{3}{5}$$

$$P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{2}$$

$$P\left( {\dfrac{A}{{A\,\,\, \cup \,\,B}}} \right) = \dfrac{5}{6}$$

$$P\left( {\dfrac{{A\,\,\, \cap \,\,B}}{{\overline A \,\,\, \cup \,\,\overline B }}} \right) = 0$$

Solution

Question 2

Occurrence of E $$\Rightarrow$$ occurrence of F

Occurrence of F $$\Rightarrow$$ occurrence of E

Non-occurrence of E $$\Rightarrow$$ non-occurrence of F

None of the above implications holds

Solution

Question 3

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

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

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

None of these

Solution

Question 4

$$4! \times 3!$$

$$4!$$

$$3! \times 3!$$

none of these

Solution

Question 5

$$[0, 1]$$

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

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

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

Solution

Question 6

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

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

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

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

Solution

Question 7

$$\dfrac{1}{18}$$

$$\dfrac{13}{18}$$

$$\dfrac{5}{18}$$

$$\dfrac{1}{9}$$

Solution

Question 8

$$1$$

$$0$$

$$0.5$$

$$-1$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$\cfrac { 10 }{ 17 } $$

$$\cfrac { 1 }{ 17 } $$

$$\cfrac { 4 }{ 17 } $$

$$\cfrac { 5 }{ 17 } $$

Solution

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