Chapter 14: Probability

Ans: According to question.

Total numbers of numbers in the draw is 20

Numbers to be selected is 6

Therefore,

\[n\left( S \right){ = ^{20}}{C_6}\]

\[n\left( S \right)=\frac{\left| \underline{20} \right.}{\left| \underline{6\left| \underline{20-6} \right.} \right.}\]

\[n\left( S \right) = 38760\]

Let  be the event that six numbers match with the six numbers already fixed by the lottery committee.

\[n\left( A \right){ = ^6}{C_6}\]

\[n\left( A \right) = 1\]

Therefore, the probability of winning the prize,

\[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

\[P\left( A \right) = \dfrac{1}{{38760}}\]

Ans:\[P\left( {A \cap B} \right) > P\left( A \right)\]

Therefore, the given probabilities are not consistently defined.

Ans: As we know that,

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

\[0.8 = 0.5 + 0.4--P(A \cap B)\]

\[0.8 - 0.9 = --P(A \cap B)\]

\[P(A \cap B) = 0.1\]

Therefore, \[P\left( {A \cap B} \right) < P\left( A \right)\] and \[P\left( {A \cap B} \right) < P\left( B \right)\]

So, the given probabilities are consistently defined.

Ans:

(i) We know that,

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

\[P(A \cup B) = \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{{15}}\]

\[P(A \cup B) = \dfrac{7}{{15}}\]

Ans:

(ii) We know that,

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

\[0.6 = 0.35 + P\left( B \right) - 0.25\]

\[P\left( B \right) = 0.6 - 0.35 + 0.25\]

\[P\left( B \right) = 0.5\]

Ans: 

(iii) We know that,

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

\[0.7 = 0.5 + 0.35 - P(A \cap B)\]

\[P(A \cap B) = 0.5 + 0.35 - 0.7\]

\[P(A \cap B) = 0.15\]

Ans: \[P\left( A \right) = \dfrac{3}{5}\] and \[P\left( B \right) = \dfrac{1}{5}\]

If A and B are mutually exclusive

\[P(A \cup B) = P\left( A \right) + P\left( B \right)\]

\[P(A \cup B) = \dfrac{3}{5} + \dfrac{1}{5}\]

\[P(A \cup B) = \dfrac{4}{5}\]

Ans: We have, \[P\left( E \right) = \dfrac{1}{4}\], \[P\left( F \right) = \dfrac{1}{2}\], \[P\left( {E \cap F} \right) = \dfrac{1}{8}\]

We know that,

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

\[P(A \cup B) = \dfrac{1}{4} + \dfrac{1}{2} - \dfrac{1}{8}\]

\[P(A \cup B) = \dfrac{5}{8}\]

Ans: \[P\left( {{\text{not E and not F}}} \right) = P\left( {\overline E  \cap \overline F } \right)\]

\[P\left( {{\text{not E and not F}}} \right) = P\left( {\overline {E \cup F} } \right)\]

\[P\left( {{\text{not E and not F}}} \right) = 1 - P\left( {E \cup F} \right)\]

\[P\left( {{\text{not E and not F}}} \right) = 1 - \dfrac{5}{8}\]

\[P\left( {{\text{not E and not F}}} \right) = \dfrac{3}{8}\]

Ans: We have,

\[P\left( {{\text{not E or not F}}} \right) = 0.25\]

\[P\left( {\overline {\text{E}}  \cup \overline {\text{F}} } \right) = 0.25\]

\[P\left( {\overline {E \cap F} } \right) = 0.25\]

\[1 - P\left( {E \cap F} \right) = 0.25\]

\[P\left( {E \cap F} \right) = 1 - 0.25\]

\[P\left( {E \cap F} \right) = 0.75\]

\[P\left( {E \cap F} \right) \ne 0\]

Therefore, E and F are not mutually exclusive event.

Ans: We know that,

\[P\left( {{\text{not A}}} \right) = 1 - P\left( A \right)\]

\[P\left( {{\text{not A}}} \right) = 1 - 0.42\]

\[P\left( {{\text{not A}}} \right) = 0.58\]

Ans: We know that,

\[P\left( {{\text{not B}}} \right) = 1 - P\left( B \right)\]

\[P\left( {{\text{not B}}} \right) = 1 - 0.48\]

\[P\left( {{\text{not B}}} \right) = 0.52\]

Ans: We know that,

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

\[P(A \cup B) = 0.42 + 0.48 - 0.16\]

\[P(A \cup B) = 0.74\]

Ans: Let A be the event that the student is studying mathematics and B be the event that the student is studying biology,

Therefore,

\[P\left( A \right) = \dfrac{{40}}{{100}}\]

\[P\left( A \right) = \dfrac{2}{5}\]

And,

\[P\left( B \right) = \dfrac{{30}}{{100}}\]

\[P\left( B \right) = \dfrac{3}{{10}}\]

So, student studying both subjects,

\[P\left( {A \cap B} \right) = \dfrac{{10}}{{100}}\]

\[P\left( {A \cap B} \right) = \dfrac{1}{{10}}\]

Therefore, probability of studying mathematics or biology is given as,

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

\[P(A \cup B) = \dfrac{2}{5} + \dfrac{3}{{10}} - \dfrac{1}{{10}}\]

\[P(A \cup B) = \dfrac{6}{{10}}\]

\[P(A \cup B) = \dfrac{3}{5}\]

Therefore, \[\dfrac{3}{5}\] is the probability that student will studying mathematics or biology.

Ans: Let probability of a randomly chosen student passing the first examination is 0.8 be \[P\left( A \right)\].

And also probability of passing the second examination is 0.7 be \[P\left( B \right)\]

Therefore,

\[P(A \cup B)\;\]is probability of passing at least one of the examination

So, We know that,

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

\[0.95 = 0.8 + 0.7 - P(A \cap B)\]

\[P(A \cap B) = 0.8 + 0.7 - 0.95\]

\[P(A \cap B) = 0.55\]

therefor, 0.55 is the probability that student will pass both the examinations.

Ans: Let probability of passing the English examination is 0.75 be \[P\left( A \right)\].

And also assume the probability of passing the Hindi examination is be \[P\left( B \right)\].

We hve given, \[P\left( A \right) = 0.75,{\text{ }}P\left( {A \cap B} \right) = 0.5,{\text{ }}P\left( {A' \cap B'} \right) = 0.1\]

We know that, 

\[P\left( {A' \cap B'} \right) = 1--P(A \cup B)\]

\[P(A \cup B) = 1--P\left( {A' \cap B'} \right)\]

\[P(A \cup B) = 1 - 0.1\]

\[P(A \cup B) = 0.9\]

We know that,

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

\[0.9 = 0.75 + P\left( B \right) - 0.5\]

\[P\left( B \right) = 0.9 + 0.5 - 0.75\]

\[P\left( B \right) = 0.65\]

Therefore, the probability of passing  the hindi examination is 0.65.