Engineering Mathematics Test 7

\(P\left( A \right) \to\) probability of getting more than one tail

\(\therefore P\left( A \right) = \frac{{11}}{{16}}\)

\(P\left( B \right) \to\) probability of getting at least one tail

\(\therefore P\left( B \right) = \frac{{15}}{{16}}\)

\(P\left( {A \cap B} \right) \to\) probability of getting more than one tail and at least one is head.

\(\therefore P\left( {A \cap B} \right) = \frac{{10}}{{16}}\) 

\(\therefore P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{{\frac{{10}}{{16}}}}{{\frac{{15}}{{16}}}}\)

A and B are mutually exhaustive events

Therefore, P(A ∪ B) = 1

 P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

\(1=\frac{3}{4}+\frac{1}{2}-P\left( A\cap B \right)\)

\(\therefore P\left( A\cap B \right)=\frac{1}{4} = 0.25\)

P(A’ ∪ B’) = P(A ∩ B)’

P(A ∩ B) + P(A ∩ B)’ = 1

\(\frac{1}{4}~+P{{\left( A\cap B \right)}^{'}}=1\)

\(P\left( {A}'\cup {B}' \right)=P{{\left( A\cap B \right)}^{'}}=\frac{3}{4}=0.75\)

Let the number of students be 100.

\(P\left( N \right) \to\) probability that students failed in Computer Networks

\(\therefore P\left( N \right) = \frac{{30}}{{100}} = \frac{3}{{10}}\)

\(P\left( A \right) \to\) probability that students failed in Computer Architecture

\(\therefore P\left( A \right) = \frac{{20}}{{100}} = \frac{2}{{10}}\)

\(P\left( {N \cap \;A} \right) \to \;{\rm{probability\;that\;the\;student\;has\;failed\;in\;both\;the\;subject}}\)

\(\therefore P\left( {N \cap \;A} \right) = \frac{{10}}{{100}} = \frac{1}{{10}}\)

\(P\left( {N\;U\;A} \right) \to \;{\rm{probability\;that\;the\;student\;has\;failed\;in\;at\;least\;one\;subject}}\)

\(P\left( {N\;U\;A} \right) = P\left( N \right) + P\left( A \right) - \;P\left( {N \cap \;A} \right)\;\) 

\(P\left( {N\;U\;A} \right) = \frac{3}{{10}} + \frac{2}{{10}}\; - \frac{1}{{10}}\) 

\(\therefore P\left( {N\;U\;A} \right) = \;\frac{4}{{10}} = \frac{2}{5}\)

Concept:

Probability: It is a measure of the likelihood of an event to occur.

Probability of an event to occur, P(E) = Favourable outcomes/Total Outcomes 

Note: If two events are independent then, the probability of occurring both of them is the product of the probability of both events.

Calculation:

For drawing the first card, we have a total 52 cards and 4 cards are ace.

So, the probability for drawing one card = \(\frac{4}{52}\)

Now, for drawing the second card without replacement. We have 51 cards and 3 ace cards.

SO, the probability for drawing the second card = \(\frac{3}{51}\)

Hence, the required probability = \(\frac{4}{52}\times \frac{3}{51}\)

=\(\frac{1}{221}\)

Total number of ways of selecting x is 100.

Now the given condition is x + (100/x) > 50. Let

\(x + \frac{{100}}{x} > 50 \Rightarrow {x^2} - 50x + 100 > 0 \Rightarrow x > 47.91\)

On analysing, we see that this equation is satisfied for all the number x such that x > 47.9 and also for x = 1 and 2

So, favourable number cases is 55.

Sample space: 7 green balls and 11 blue balls

Probability that 1^st one is green = \(\frac{7}{{7 + 11}} = \frac{7}{{18}}\)

Since green ball is not replaced

sample space:  6 green balls and 11 blue balls

Probability that 2^nd one is green = \(\frac{6}{{6 + 11}} = \frac{6}{{17}}\)

Since green ball is not replaced

sample space:  5 green balls and 11 blue balls

Probability that 3^rd one is green = \(\frac{5}{{5 + 11}} = \frac{5}{{16}}\)

probability for all three balls being green = \(\frac{7}{{18}} \times \frac{6}{{17}} \times \frac{5}{{16}} = \frac{{35}}{{816}}\;\)

Tips and Tricks:

Data:  

\(p\left( P \right) = \frac{1}{4}\)

\(P\left( {\frac{P}{Q}} \right) = \;\frac{1}{2}\) , \(P\left( {\frac{Q}{P}} \right) = \;\frac{1}{3}\)

Formula

\(P\left( {\frac{A}{B}} \right) = \;\frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\)

Calculation:

\(P\left( {\frac{Q}{P}} \right) = \;\frac{{P\;\left( {P \cap Q} \right)}}{{P\left( P \right)}},\;\;\)

\(\frac{1}{3} = \;\frac{{P\left( {P \cap Q} \right)}}{{\frac{1}{4}}}\)  , 

\(P\left( {P \cap Q} \right) = \;\frac{1}{{12}}\)

Also, \(P\left( {\frac{P}{Q}} \right) = \frac{{P\;\left( {P \cap Q} \right)}}{{P\left( Q \right)}},\;\;\)

\(\frac{1}{2} = \frac{{\frac{1}{{12}}}}{{P\left( Q \right)}}\) ,

\(P\left( Q \right) = \frac{1}{6}\)

Required probability, \(P\left( {\frac{{P'}}{{Q'}}} \right) = \frac{{P\left( {P' \cap Q'} \right)}}{{P\left( {Q'} \right)}}\)

\(= \frac{{P{{\left( {P \cup Q} \right)}'}}}{{1 - P\left( Q \right)}}\; = \frac{{1 - P\left( {P \cup Q} \right)}}{{1 - P\left( Q \right)}}\)

\(= \frac{{1 - \left( {P\left( P \right) + P\left( Q \right) - P\;\left( {P \cap Q} \right)} \right)}}{{1 - P\left( Q \right)}}\;\)

\(= \frac{{1 - \left( {\frac{1}{4} + \frac{1}{6} - \frac{1}{{12}}} \right)}}{{1 - \frac{1}{6}}}\)

Explanation:

Let, T: A speaks truth ⇒ P(T) = ¾

T̅ : A lies ⇒ P(T̅) = 1 - P(T) = ¼

Let, B: Shyam won the match.

There are three cases for matches. It can be won, drawn or lost.

The probability of winning a match, P (B/T) = 1/3

The probability of not winning a match, P(B,T̅) = 2/3

Using Baye’s theorem:

\(P\left( {\frac{T}{B}} \right) = \frac{{P\left( T \right).P\left( {\frac{B}{T}} \right)}}{{P\left( T \right).P\left( {\frac{B}{T}} \right) + P\left( {\bar T} \right).P\left( {\frac{B}{{\bar T}}} \right)}} \)

\(P(\frac{T}{B}) = \frac{{\frac{3}{4} \times \frac{1}{3}}}{{\frac{3}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{2}{3}}} = \frac{3}{5} = 0.6\)

Bag A → 20% red balls, 80% green balls

Bag B → 30% red balls, 60% green balls, 10% yellow balls

As both the bags have equal number of balls,

After mixing of both the bags,

% of red balls \(= \frac{{20 + 35}}{2} = 25\% \)

% green balls \(= \frac{{80 + 60}}{2} = 70\%\)

% yellow balls \(= \frac{{10}}{2} = 5\% \)

The probability to get a red ball \(= \frac{{25}}{{100}} \times 100\)

Data:

\(P\left( A \right) = \frac{7}{{10}} = 0.7,\;P\left( {Y/A} \right) = 0.6\)

\(P\left( B \right) = \frac{3}{{10}} = 0.3,\;P\left( {Y/B} \right)\; = 0.3\)

Formula:

P(Y) = P(A) × P(Y/A) + P(B) × P(Y/B)

Calculation:

P(Y) = 0.7 × 0.6 + 0.3 × 0.3

P(Y) = 0.51