Chapter 14: Probability

Ans. Total number of marbles \[ = 60\]

The number of ways of drawing \[5\] marbles from \[60\] marbles is \[^{60}{C_5}\] .

All the marbles drawn will be blue if we draw \[5\] marbles out of the \[20\] blue marbles. \[5\] blue marbles can be drawn out of the \[20\] blue marbles in \[^{20}{C_5}\] ways.

Probability that all marbles will be blue is \[\frac{{^{20}{C_5}}}{{^{60}{C_5}}}\] .

Ans. Total number of marbles \[ = 60\]

Number of ways of drawing \[5\] marbles from \[60\] marbles is \[^{60}{C_5}\] .

The number of ways in which the drawn marbles is not green, that is, when red or blue marbles are drawn, is \[^{20 + 10}{C_5}{ = ^{30}}{C_5}\].

So, probability that no marble is green is \[\frac{{^{30}{C_5}}}{{^{60}{C_5}}}\] .

Therefore, the probability that at least one marble will be green is \[1 - \frac{{^{30}{C_5}}}{{^{60}{C_5}}}\].

Ans. Number of ways of drawing \[4\] cards from a pack of \[52\] cards are \[^{52}{C_4}\] .

In a deck, there are \[13\] diamonds and \[13\] spades.

So, the number of ways of drawing three diamonds and one spade is \[^{13}{C_3}{ \times ^{13}}{C_1}\].

Hence, the probability of drawing three diamonds and one spade is \[\frac{{^{13}{C_3}{ \times ^{13}}{C_1}}}{{^{52}{C_4}}}\].

Ans. Total number of faces of the die is \[6\].

The number of faces with the number ‘ \[2\] ’ is \[3\].

Therefore, probability of obtaining the number \[2\] is \[P(2) = \frac{{n(2)}}{{n(S)}}\] .

\[ \Rightarrow P(2) = \frac{3}{6}\]

\[ \Rightarrow P(2) = \frac{1}{2}\]

Ans. Total number of faces of the die is \[6\].

Number of faces with the number ‘ \[1\] ’ is \[2\] and with the number ‘ \[3\] ’ is \[1\].

Therefore, probability of obtaining the number \[2\] or \[3\] is \[P(1 \cup 3) = \frac{{n(1 \cup 3)}}{{n(S)}}\] .

\[ \Rightarrow P(1 \cup 3) = \frac{{1 + 2}}{6}\] 

\[ \Rightarrow P(1 \cup 3) = \frac{1}{2}\]

Ans. Total number of faces of the die is \[6\].

Number of faces with the number ‘ \[3\] ’ is \[1\].

Therefore, probability of obtaining the number \[3\] is \[P(3) = \frac{1}{6}\] .

\[ \Rightarrow P(3') = 1 - P(3)\] 

\[ \Rightarrow P(3') = 1 - \frac{1}{6}\]

\[ \Rightarrow P(3') = \frac{5}{6}\]

Ans. Total number of tickets sold is \[10,000\].

Number of prizes awarded is \[10\].

If A is the event we buy \[1\] ticket, the probability of winning will be \[P(A) = \frac{{n(A)}}{{n(S)}}\] .

\[ \Rightarrow P(A) = \frac{{10}}{{10,000}}\]

\[ \Rightarrow P(A) = \frac{1}{{1,000}}\] 

So, the probability of not getting a prize is \[P(A') = 1 - P(A)\].

\[ \Rightarrow P(A') = 1 - \frac{1}{{1,000}}\]

\[ \Rightarrow P(A') = \frac{{999}}{{1,000}}\]

Ans. Total number of tickets sold is \[10,000\].

Number of prizes awarded is \[10\].

Number of tickets on which a prize is not awarded is \[10,000 - 10 = 9,990\].

If B is the event we buy \[2\] tickets, the probability of not winning will be \[P(B) = \frac{{^{9,990}{C_2}}}{{^{10,000}{C_2}}}\] .

Ans. Total number of tickets sold is \[10,000\].

Number of prizes awarded is \[10\].

Number of tickets on which a prize is not awarded is \[10,000 - 10 = 9,990\].

If C is the event we buy \[10\] tickets, the probability of not winning will be \[P(C) = \frac{{^{9,990}{C_{10}}}}{{^{10,000}{C_{10}}}}\] .

Ans. You and your friend are among the \[100\] students. 

Total number of ways of selecting \[2\] students out of \[100\] students is ¹⁰⁰C₂.

If both of you enter the same section, you are either among the section having \[40\] students or the one having \[60\] students.

The number of ways in which you both can be in the same section is \[^{40}{C_2}{ + ^{60}}{C_2}\]. 

The probability that you both can be in the same section is \[\frac{{^{40}{C_2}{ + ^{60}}{C_2}}}{{^{100}{C_2}}}\]

\[ \Rightarrow \frac{{\frac{{40!}}{{2!38!}} + \frac{{60!}}{{2!58!}}}}{{\frac{{100!}}{{2!98!}}}}\]

\[ \Rightarrow \frac{{(40 \times 39) + (60 \times 59)}}{{(100 \times 99)}}\]

\[ \Rightarrow \frac{{17}}{{33}}\]

Ans. The probability that you both can be in the same section is \[\frac{{17}}{{33}}\].

The probability that you both are not in the same section will be \[1 - \frac{{17}}{{33}}\].

\[ \Rightarrow \frac{{16}}{{33}}\]

Ans. Let \[{L_1},{L_2},{L_3}\] be the three letters and \[{E_1},{E_2},{E_3}\] be their corresponding envelopes.

There are \[6\] ways of inserting \[3\] letters in \[3\] envelopes. These are as follows:

\[\begin{array}{l}{L_1}{E_1},{L_2}{E_3},{L_3}{E_2}\\{L_2}{E_2},{L_1}{E_3},{L_3}{E_1}\\{L_3}{E_3},{L_1}{E_2},{L_2}{E_1}\\{L_1}{E_1},{L_2}{E_2},{L_3}{E_3}\\\;{L_1}{E_2},{L_2}{E_3},{L_3}{E_1}\\{L_1}{E_3},{L_2}{E_1},{L_3}{E_2}\end{array}\]

We observe that there are \[4\] ways in which at least one letter is inserted in its proper envelope.

Hence, the required probability is \[\frac{4}{6} = \frac{1}{2}\] .

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

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

\[ \Rightarrow P(A \cup B) = 0.54 + 0.69 - 0.35\]

\[ \Rightarrow P(A \cup B) = 0.88\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P\left( {A' \cap B'} \right) = P(A \cup B)'\]

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

\[ \Rightarrow P\left( {A' \cap B'} \right) = 1 - 0.88\]

\[ \Rightarrow P\left( {A' \cap B'} \right) = 0.12\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P\left( {A \cap B'} \right) = P\left( A \right)--P\left( {A \cap B} \right)\]

\[ \Rightarrow P\left( {A \cap B'} \right) = 0.54--0.35\]

\[ \Rightarrow P\left( {A \cap B'} \right) = 0.19\]

Ans. It is given to us that \[P(A) = 0.54\] , \[P(B) = 0.69\] , \[P(A \cap B) = 0.35\] .

We know that \[P\left( {B \cap A'} \right) = P\left( B \right)--P\left( {A \cap B} \right)\]

\[ \Rightarrow P\left( {B \cap A'} \right) = 0.69--0.35\]

\[ \Rightarrow P\left( {B \cap A'} \right) = 0.34\]

Ans. Let E be the event that the spokesperson will be a male and F be the event that the spokesperson will be over \[35\] years of age.

Subsequently, \[P(E) = \frac{3}{5}\] and \[P(F) = \frac{2}{5}\] .

Since there is only one male who is above  \[35\] years of age, \[P(E \cap F) = \frac{1}{5}\] .

We know that \[P(E \cup F) = P(E) + P(F) - P(E \cap F)\]

\[ \Rightarrow P(E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5}\] 

\[ \Rightarrow P(E \cup F) = \frac{4}{5}\] 

Hence, the probability that the spokesperson will either be a male or over \[35\] years of age is \[\frac{4}{5}\].

Ans. Since four-digit numbers greater than \[5,000\] are formed, the thousand’s digit is either \[5\] or \[7\] . The remaining three digits can be filled by any of the given digits with repetition. 

So, the total number of ways of forming four-digit numbers are \[2 \times 5 \times 5 \times 5\], and subtract \[1\] from the obtained answer, as we cannot count \[5,000\] in the numbers.

\[ \Rightarrow 250 - 1\]

\[ \Rightarrow 249\] ways

A number is divisible by \[5\] if the digit at its unit’s place is either \[0\] or \[5\] .

Total number of four-digit   numbers   greater than \[5,000\]   that   are   divisible by \[5\] are \[(2 \times 5 \times 5 \times 2) - 1\] .

\[ \Rightarrow 100 - 1\]

\[ \Rightarrow 99\]

Therefore, the probability of forming a number divisible by \[5\] when the digits are repeated is \[\frac{{99}}{{249}} = \frac{{33}}{{83}}\]

Ans. Since four-digit numbers greater than \[5,000\] are formed, the thousand’s digit is either \[5\] or \[7\] . The remaining three digits can be filled by any of the given digits without repetition.

So, the total number of ways of forming four-digit numbers is \[2 \times 4 \times 3 \times 2 = 48\].

A number is divisible by \[5\] if the digit at its unit’s place is either \[0\] or \[5\].

When the digit at the thousands place is \[5\], the unit place can be filled only with \[0\], so the number of possible ways is \[1 \times 3 \times 2 \times 1 = 6\].

When the digit at the thousands place is \[7\] , the units place can be filled only with \[0\] or \[5\] , so the number of possible ways are \[1 \times 3 \times 2 \times 2 = 12\] .

Total number of four-digit   numbers   greater than \[5,000\]   that   are   divisible by \[5\] are \[12 + 6 = 18\] .

Therefore, the probability of forming a number divisible by \[5\] when the digits are repeated is \[\frac{{18}}{{48}} = \frac{3}{8}\] .

Ans. The number lock of a suitcase has \[4\] wheels, each of which is labelled with ten digits, that is, from \[0\] to \[9\].

Number of ways of selecting \[4\] different digits out of \[10\] digits is \[^{10}{C_4}\] .

Further, each combination of \[4\] different digits can be arranged in \[4\] ways.

Number of four digits with no repetitions \[{ = ^{10}}{C_4} \times 4!\]

\[ \Rightarrow \frac{{10!}}{{4!6!}} \times 4!\]

\[ \Rightarrow 10 \times 9 \times 8 \times 7\]

\[ \Rightarrow 5040\]

There is only one number that can open the suitcase.

Thus, the required probability is \[\frac{1}{{5040}}\] .