Chapter 16: Probability

Ans: Given: Student will pass his examination\[ = 0.73\]

Student getting a compartment \[ = 0.13\]

Student will either pass or get compartment \[0.96.\]

The probability,

$P(A) = 0.73,$

$P(B) = 0.13$ 

\[P(AUB) = 0.96\]

Find \[P(A \cap B)\]

By General addition rule,

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Substitute the values to general addition rule,

\[ \Rightarrow {\text{P}}({\text{A}} \cup {\text{B}}) = 0.73 + 0.13 - 0\]

\[\therefore A\] and \[B\] are independent events 

\[ \Rightarrow {\text{P}}({\text{A}} \cup {\text{B}}) = 0.86\]

But \[P(A \cup B) = 0.96\]

Hence, the given statement is false.

Ans: Given: Probabilities that a typist will make \[0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\]

Mistakes in typing a report are\[,{\text{ }}0.12,{\text{ }}0.25,{\text{ }}0.36,{\text{ }}0.14,{\text{ }}0.08,{\text{ }}0.11.\]

Sum of all probabilities \[ = 1\]

$\therefore P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11$ 

\[ = 1.06 > 1\]

Hence, the given statement is False. 

Ans: Given: \[{\text{P}}({\text{A}}) = 0.5\] and

\[P(A \cap B) \leqslant 0.3\]

The probability,

\[{\text{P}}({\text{A}}) = 0.5\]

\[P(A \cap B) \leqslant 0.3\]

\[P(A) \times P(B) \leqslant 0.3\]

Substitute,

\[ \Rightarrow 0.5 \times P(B) \leqslant 0.3\]

\[ \Rightarrow P(B) \leqslant \dfrac{{0.3}}{{0.5}}\]

\[ \Rightarrow P(B) \leqslant 0.6\]

Hence the given statement is False.

Ans: Given:The probability of intersection of two events\[A\] and \[B\]

\[A\] and \[B\] are two events,

Here,

\[A \cap B \subset A\]

\[ \Rightarrow P(A \cap B) \leqslant P(A)\]

This is always true.

Hence, the given statement is true.

Ans: Given: \[P(A)\]\[ = .7\]

\[P(B) = .3\]

The probability of an occurrence of both \[ = .4\]

The probability,

  ${\text{P}}({\text{A}}) = 0.7$

  ${\text{P}}({\text{B}}) = 0.3$ 

\[\therefore P(A \cap B) = P(A) \times P(B)\]

Substitute the values of\[P(A)\] and \[P(B)\],

  $P(A \cap B) = 0.7 \times 0.3$

   $= 0.21$ 

But the given probability is \[0.4\].

Hence, the given statement is False.

Ans: Given: Distinction in their final examinations \[ = 1.2.\]

The probability of each student getting distinction in their final examination is less than or equal to \[1\]

Since, the two given events are not related to the given sample space.

Therefore,

The sum of the probabilities of two may be \[1.2\].

Hence, the given statement is true.

Ans: Given:

  $P(A) = 0.77$

  $P(B) = 0.08$ 

The probability that it will tie the game \[ = 0.08\],

\[A\] be the event that home team will win the football game

\[B\] be the event that home team will tie the game and,

\[C\] be the event that they will lose the game

Then,

$P(A) = 0.77$

$P(B) = 0.08$ 

Find the\[P\] or \[P(C)\]

Sum of all Probabilities\[ = 1\]

\[\therefore {\text{P}}({\text{A}}) + {\text{P}}({\text{B}}) + {\text{P}}({\text{C}}) = 1\]

Substitute the values,

\[ \Rightarrow 0.77 + 0.08 + {\text{P}}({\text{C}}) = 1\]

\[ \Rightarrow {\text{P}}({\text{C}}) = 1 - (0.77 + 0.08)\]

\[P(C)\]\[ = 0.15\].

The probability that the home team will win an upcoming football game is \[0.77\], the probability that it will tie the game is\[0.08\], and the probability that it will lose the game is \[0.5.\]

Ans: Given: Four elementary outcomes\[ = e_1,{\text{ }}e_2,{\text{ }}e_3,{\text{ }}e_4\]

  $P\left( {e_1{\text{ }}} \right) = .1,{\text{ }}$

  $P\left( {e_2{\text{ }}} \right) = .5,{\text{ }}$

  $P\left( {e_3{\text{ }}} \right) = .1$ 

Sum of all probabilities\[ = 1\]

\[\therefore {\text{P}}\left( {{{\text{e}}_1}} \right) + {\text{P}}\left( {{{\text{e}}_2}} \right) + {\text{P}}\left( {{{\text{e}}_3}} \right) + {\text{P}}\left( {{{\text{e}}_4}} \right) = 1\]

Substitute the values,

\[ \Rightarrow 0.1 + 0.5 + 0.1 + {\text{P}}\left( {{{\text{e}}_4}} \right) = 1\]

\[ \Rightarrow {\text{P}}\left( {{{\text{e}}_4}} \right) = 1 - 0.7\]

\[ \Rightarrow {\text{P}}\left( {{{\text{e}}_4}} \right) = 0.3\]

Therefore ,

If \[e_1,{\text{ }}e_2,{\text{ }}e_3,{\text{ }}e_4\] are the four elementary outcomes in a sample space \[P\left( {e_1{\text{ }}} \right) = .1,{\text{ }}P\left( {e_2{\text{ }}} \right) = .5,{\text{ }}P\left( {e_3{\text{ }}} \right) = .1,\]then the probability of \[e_4\] is \[0.3.\]

Ans: Given: \[S{\text{ }} = {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\}\]

\[E{\text{ }} = {\text{ }}\left\{ {1,{\text{ }}3,{\text{ }}5} \right\},\]

Given that,

\[{\text{S}} = \{ 1,2,3,4,5,6\} \]

\[{\text{E}} = \{ 1,3,5\} \]

\[\overline {\text{E}}  = {\text{S}} - {\text{E}}\]

\[ = \{ 1,2,3,4,5,6\} \]\[ - \{ 1,3,5\} \]

\[ = \{ 2,4,6\} \]

Hence, the value of the filler is \[\left\{ {2,4,6} \right\}.\]

Therefore,

Let \[S = \left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\}\] and \[{\text{E}} = {\text{ }}\left\{ {1,{\text{ }}3,{\text{ }}5} \right\},\]then \[\overline {\text{E}} \] is \[\left\{ {2,4,6} \right\}.\]

Ans: Given:

 ${\text{P}}({\text{A}}) = 0.3,$

 $ {\text{P}}({\text{B}}) = 0.2$ 

\[{\text{P}}({\text{A}} \cap {\text{B}}) = 0.1,\]

Given that,

  ${\text{P}}({\text{A}}) = 0.3,$

  ${\text{P}}({\text{B}}) = 0.2$ 

\[{\text{P}}({\text{A}} \cap {\text{B}}) = 0.1\]

\[{\text{P}}({\text{A}} \cap \overline{{B}} ) = {\text{P}}({\text{A}}) - {\text{P}}({\text{A}} \cap {\text{B}})\]

Substitute the values,\[P(A)\] and \[P(B)\]

\[{\text{P}}({\text{A}} \cap {\text{B}})\]\[ = 0.3 - 0.1\]

\[ = 0.2\]

Hence, the value of the filler is \[0.2.\]

Therefore the statement is,

If \[{\text{A}}\] and \[{\text{B}}\] are two events associated with a random experiment such that \[P(A) = 0.3,{\text{P}}({\text{B}}) = 0.2\] and \[{\text{P}}({\text{A}} \cap {\text{B}}) = 0.1,\] then the value of \[{\text{P}}({\text{A}} \cap \overline{{B}} )\] is \[0.2\].