Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Let A, B and C be three events, which are pair-wise independent and $$\overrightarrow E $$ denotes the completement of an event E. If $$P\left( {A \cap B \cap C} \right) = 0$$ and $$P\left( C \right) > 0,$$ then $$P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]$$ is equal to :

A

$$P\left( {\overline A } \right) - P\left( B \right)$$

B

$$P\left( A \right) + P\left( {\overline B } \right)$$

C

$$P\left( {\overline A } \right) - P\left( {\overline B } \right)$$

D

$$P\left( {\overline A } \right) + P\left( {\overline B } \right)$$

Here, $$P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}$$

= $${{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}$$

= $${{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}$$

= $${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}$$

= $${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}$$ ($$ \because $$$$\left. {P\left( {A \cap B \cap C} \right) = 0} \right)$$

= $${{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}$$

[$$ \because $$ A, B and C are independent events]

= 1 - P(A) - P(B)

= $$P\left( {\overline A } \right)$$ - P(B) or $$P\left( {\overline B } \right)$$ - P(A)

= $${{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}$$

= $${{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}$$

= $${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}$$

= $${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}$$ ($$ \because $$$$\left. {P\left( {A \cap B \cap C} \right) = 0} \right)$$

= $${{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}$$

[$$ \because $$ A, B and C are independent events]

= 1 - P(A) - P(B)

= $$P\left( {\overline A } \right)$$ - P(B) or $$P\left( {\overline B } \right)$$ - P(A)

2

Two different families A and B are blessed with equal numbe of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $${1 \over {12}},$$ then the number of children in each family is :

A

3

B

4

C

5

D

6

Let the number of children in each family be x.

Thus the total number of children in both the families are 2x

Now, it is given that 3 tickets are distributed amongst the children of these two families.

Thus, the probability that all the three tickets go to the children in family B

= $${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$$ = $${1 \over {12}}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$$ = $${1 \over {12}}$$

$$ \Rightarrow $$ $${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$$ = $${1 \over 6}$$

Thus, the number of children in each family is 5.

Thus the total number of children in both the families are 2x

Now, it is given that 3 tickets are distributed amongst the children of these two families.

Thus, the probability that all the three tickets go to the children in family B

= $${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$$ = $${1 \over {12}}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$$ = $${1 \over {12}}$$

$$ \Rightarrow $$ $${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$$ = $${1 \over 6}$$

Thus, the number of children in each family is 5.

3

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P (X = 2) equals :

A

$$25 \over 169$$

B

$$49\over 169$$

C

$$24 \over 169$$

D

$$52 \over 169$$

P (X = 1) means out of two drawn cards one card is ace.

and P(X = 2) means both the drawn cards are ace.

$$ \therefore $$ P(X = 1) = first card is ace or 2nd card is ace.

= A $$-$$ $$+$$ $$-$$ A

= $${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$$

= $$2 \times {4 \over {52}} \times {{48} \over {52}}$$

P(X = 2) =First and second both cards arc ace.

= A A

= $${4 \over {52}} \times {4 \over {52}}$$

$$ \therefore $$ P(X = 1) + P(X = 2)

= $$2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}$$

= $${{25} \over {169}}$$

and P(X = 2) means both the drawn cards are ace.

$$ \therefore $$ P(X = 1) = first card is ace or 2nd card is ace.

= A $$-$$ $$+$$ $$-$$ A

= $${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$$

= $$2 \times {4 \over {52}} \times {{48} \over {52}}$$

P(X = 2) =First and second both cards arc ace.

= A A

= $${4 \over {52}} \times {4 \over {52}}$$

$$ \therefore $$ P(X = 1) + P(X = 2)

= $$2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}$$

= $${{25} \over {169}}$$

4

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :

A

$${{21} \over {49}}$$

B

$${{27} \over {49}}$$

C

$${{26} \over {49}}$$

D

$${{32} \over {49}}$$

5 Red and 2 green balls

P(one red ball) = $${5 \over 7}$$

P(one green ball) = $${2 \over 7}$$

**Case I : **

If drawn ball is green than a red ball is added

$$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$$ P (red ball) = $${6 \over 7}$$

**Case II : **

If drawn ball is red than a green ball is added

$$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$$ P (red ball) = $${4 \over 7}$$

P (2^{nd} red ball) = $${5 \over 7}$$ $$ \times {4 \over 7} + {2 \over 7} \times {6 \over 7}$$ = $${{32} \over {49}}$$

P(one red ball) = $${5 \over 7}$$

P(one green ball) = $${2 \over 7}$$

If drawn ball is green than a red ball is added

$$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$$ P (red ball) = $${6 \over 7}$$

If drawn ball is red than a green ball is added

$$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$$ P (red ball) = $${4 \over 7}$$

P (2

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