Joint Entrance Examination

Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

If A and B are any two events such that P(A) = $${2 \over 5}$$ and P (A $$ \cap $$ B) = $${3 \over {20}}$$, hen the conditional probability, P(A $$\left| {} \right.$$(A' $$ \cup $$ B')), where A9 denotes the complement of A, is equal to :

A

$${1 \over 4}$$

B

$${5 \over 17}$$

C

$${8 \over 17}$$

D

$${11 \over 20}$$

$$P\left( {{A \over {A' \cup B'}}} \right)$$

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

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

$$\left[ \, \right.$$As $$\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$$

$$ = {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}$$

As $$A \cap A' = \phi $$

$$ \therefore $$ $$\phi \cup \left( {A \cap B'} \right) = A \cap B'$$

$$ = {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$$

$$\left[ \, \right.$$As $$A \cap B' = A - \left( {A \cap B} \right)$$

and $$\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$$

Given $$P\left( A \right) = {2 \over 5}$$

and $$P\left( {A \cap B} \right) = {3 \over {20}}$$

$$ = {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$$

$$ = {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$$

$$ = {5 \over {17}}$$

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

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

$$\left[ \, \right.$$As $$\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$$

$$ = {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}$$

As $$A \cap A' = \phi $$

$$ \therefore $$ $$\phi \cup \left( {A \cap B'} \right) = A \cap B'$$

$$ = {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$$

$$\left[ \, \right.$$As $$A \cap B' = A - \left( {A \cap B} \right)$$

and $$\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$$

Given $$P\left( A \right) = {2 \over 5}$$

and $$P\left( {A \cap B} \right) = {3 \over {20}}$$

$$ = {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$$

$$ = {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$$

$$ = {5 \over {17}}$$

2

An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is :

A

$${{240} \over {729}}$$

B

$${{192} \over {729}}$$

C

$${{256} \over {729}}$$

D

$${{496} \over {729}}$$

Probability of fail, P(F) = p

$$ \therefore $$ Probability of success, P(S) = 2p

We know,

p + 2Pp = 1

$$ \Rightarrow $$ p = $${1 \over 3}$$

$$ \therefore $$ Now, probability of at least 5 success in 6 trials,

P(x $$ \ge $$ 5)

= P(x = 5) + P (x = 6)

=^{6}C_{5} $${\left( {{2 \over 3}} \right)^5}{\left( {{1 \over 3}} \right)^1}{ + ^6}{C_6}{\left( {{2 \over 3}} \right)^6}{\left( {{1 \over 3}} \right)^o}$$

= $${\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)$$

= $${{256} \over {729}}$$

$$ \therefore $$ Probability of success, P(S) = 2p

We know,

p + 2Pp = 1

$$ \Rightarrow $$ p = $${1 \over 3}$$

$$ \therefore $$ Now, probability of at least 5 success in 6 trials,

P(x $$ \ge $$ 5)

= P(x = 5) + P (x = 6)

=

= $${\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)$$

= $${{256} \over {729}}$$

3

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with
replacement, then the variance of the number of green balls drawn is:

A

6

B

4

C

$${6 \over {25}}$$

D

$${{12} \over 5}$$

We can apply binomial probability distribution

n = 10

p = Probability of drawing a green ball = $${{15} \over {25}}$$ = $${3 \over 5}$$

Also q = 1 - $${3 \over 5}$$ = $${2 \over 5}$$

Variance = npq

= $$10 \times {3 \over 5} \times {2 \over 5}$$ = $${{12} \over 5}$$

n = 10

p = Probability of drawing a green ball = $${{15} \over {25}}$$ = $${3 \over 5}$$

Also q = 1 - $${3 \over 5}$$ = $${2 \over 5}$$

Variance = npq

= $$10 \times {3 \over 5} \times {2 \over 5}$$ = $${{12} \over 5}$$

4

If two different numbers are taken from the set {0, 1, 2, 3, ........, 10}; then the probability that their sum as
well as absolute difference are both multiple of 4, is:

A

$${{12} \over {55}}$$

B

$${{14} \over {45}}$$

C

$${{7} \over {55}}$$

D

$${{6} \over {55}}$$

Let A = {0, 1, 2, 3, 4, ......., 10}

Total number of ways of selecting 2 different numbers from A is

n (S) =^{11}C_{2} = 55, where 'S' denotes sample space

Let E be the given event

$$ \therefore $$ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

$$ \Rightarrow $$ n (E) = 6

$$ \therefore $$ P(E) = $${{n\left( E \right)} \over {n\left( S \right)}}$$ = $${6 \over {55}}$$

Total number of ways of selecting 2 different numbers from A is

n (S) =

Let E be the given event

$$ \therefore $$ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

$$ \Rightarrow $$ n (E) = 6

$$ \therefore $$ P(E) = $${{n\left( E \right)} \over {n\left( S \right)}}$$ = $${6 \over {55}}$$

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