Q1. Which of the following can not be valid assignment of probabilities for outcomes of sample Space \(S = \omega_1,\;\omega_2,\;\omega_3\;\omega_4,\;\omega_5,\;\omega_6,\;\omega_7,\;\) \[ \begin{aligned} \begin{array}{} \text{Assignments} &\omega_1&\omega_2&\omega_3&\omega_4&\omega_5&\omega_6&\omega_7\\ (a)&0.1&0.01&0.05&0.03&0.01&0.2&0.6\\ (b)&\frac{1}{7}&\frac{1}{7}&\frac{1}{7}&\frac{1}{7}&\frac{1}{7}&\frac{1}{7}&\frac{1}{7}\\ (c)&0.1&0.2&0.3&0.4&0.5&0.6&0.7\\ (d)&-0.1&0.2&0.3&0.4&-0.2&0.1&0.3\\ (e)&\frac{1}{14}&\frac{2}{14}&\frac{3}{14}&\frac{4}{14}&\frac{5}{14}&\frac{}{14}&\frac{15}{14}\\ \end{array} \end{aligned} \]

Solution

For any assignment to be a valid probability distribution on the sample space \(S=\{\omega_1,\omega_2,\omega_3,\omega_4,\omega_5,\omega_6,\omega_7\}\), two fundamental conditions must be satisfied: every probability must be non-negative and the total sum of all probabilities must be exactly \(1\).

We examine each case using these criteria.

For option (a),

\[ \begin{aligned} 0.1+0.01+0.05+0.03+0.01+0.2+0.6 &=0.20+0.2+0.6\\ &=1. \end{aligned} \]

All values are non-negative and the sum is \(1\); hence (a) is valid.

For option (b),

\[ \begin{aligned} 7\times \frac{1}{7}=1. \end{aligned} \]

Again, each probability is positive and the total is \(1\); therefore (b) is valid.

For option (c),

\[ \begin{aligned} 0.1+0.2+0.3+0.4+0.5+0.6+0.7 &=2.8. \end{aligned} \]

Since the sum exceeds \(1\), option (c) cannot represent a probability distribution.

For option (d), two probabilities are negative (\(-0.1\) and \(-0.2\)), which directly violates the non-negativity requirement. Hence (d) is also invalid.

For option (e), one entry is missing, and moreover \(\frac{15}{14}>1\), which alone is impossible for a probability. Therefore (e) is not a valid assignment.

Hence, the assignments which cannot be valid are (c), (d), and (e).

Q2. A coin is tossed twice, what is the probability that atleast one tail occurs?

Solution

When a coin is tossed twice, all outcomes are assumed to be equally likely. The sample space is \(S=\{HH, HT, TH, TT\}\), where \(H\) denotes head and \(T\) denotes tail.

Let \(A\) be the event that at least one tail occurs. Then,

\[ \begin{aligned} A &= \{HT,\,TH,\,TT\} \end{aligned} \]

The total number of possible outcomes is \(n(S)=4\), and the number of favourable outcomes is \(n(A)=3\).

\[ \begin{aligned} P(A) &= \frac{n(A)}{n(S)}\\ &= \frac{3}{4} \end{aligned} \]

Alternatively, this can also be verified using the complement method. The only outcome with no tail is \(HH\). Hence, if \(A'\) denotes the event “no tail occurs”, then \(A'=\{HH\}\).

\[ \begin{aligned} P(A) &= 1-P(A')\\ &= 1-\frac{1}{4}\\ &= \frac{3}{4} \end{aligned} \]

Therefore, the probability that at least one tail occurs when a coin is tossed twice is \(\frac{3}{4}\).

Q3. A die is thrown, find the probability of following events:
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.

Solution

When a die is thrown once, the sample space is \(S=\{1,2,3,4,5,6\}\). Since all outcomes are equally likely, each outcome has probability \(\frac{1}{6}\).

Let us evaluate each event separately.

(i) Prime numbers on a die are \(2,3,5\). Hence,

\[ \begin{aligned} A &= \{2,3,5\}\\ P(A) &= \frac{3}{6}=\frac{1}{2} \end{aligned} \]

(ii) Numbers greater than or equal to \(3\) are \(3,4,5,6\). Therefore,

\[ \begin{aligned} B &= \{3,4,5,6\}\\ P(B) &= \frac{4}{6}=\frac{2}{3} \end{aligned} \]

(iii) Numbers less than or equal to \(1\) include only \(1\). Hence,

\[ \begin{aligned} C &= \{1\}\\ P(C) &= \frac{1}{6} \end{aligned} \]

(iv) There is no number on a die that is more than \(6\). Thus the event is impossible.

\[ \begin{aligned} D &= \varnothing\\ P(D) &= 0 \end{aligned} \]

(v) Numbers less than \(6\) are \(1,2,3,4,5\). Hence,

\[ \begin{aligned} E &= \{1,2,3,4,5\}\\ P(E) &= \frac{5}{6} \end{aligned} \]

Therefore, the required probabilities are \(\frac{1}{2},\ \frac{2}{3},\ \frac{1}{6},\ 0,\) and \(\frac{5}{6}\) respectively.

Q4. A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is (i) an ace (ii) black card.

Solution

A standard pack contains \(52\) distinct playing cards, and each card is equally likely to be selected. Hence, the sample space consists of all \(52\) cards.

Therefore, for part (a), the number of points in the sample space is

\[ \begin{aligned} n(S) &= 52 \end{aligned} \]

For part (b), there is exactly one ace of spades in the entire deck. Since every card has equal chance of being drawn,

\[ \begin{aligned} P(\text{ace of spades}) &= \frac{1}{52} \end{aligned} \]

For part (c)(i), there are four aces in a pack of cards, one from each suit. Hence the favourable outcomes are \(4\).

\[ \begin{aligned} P(\text{ace}) &= \frac{4}{52}=\frac{1}{13} \end{aligned} \]

For part (c)(ii), black cards consist of all spades and clubs. Each suit contains \(13\) cards, so the total number of black cards is \(26\).

\[ \begin{aligned} P(\text{black card}) &= \frac{26}{52}=\frac{1}{2} \end{aligned} \]

Hence, the sample space has \(52\) points, the probability of drawing the ace of spades is \(\frac{1}{52}\), the probability of drawing an ace is \(\frac{1}{13}\), and the probability of drawing a black card is \(\frac{1}{2}\).

Q5. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is
(i) 3
(ii) 12

Solution

The coin has two faces marked \(1\) and \(6\), and the die has outcomes \(\{1,2,3,4,5,6\}\). Since both are fair, all possible ordered pairs are equally likely. Hence, the sample space consists of \(2\times 6=12\) outcomes:

\[ \begin{aligned} S=\{&(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \]

Each outcome has probability \(\frac{1}{12}\).

(i) For the sum to be \(3\), the only possible outcome is \((1,2)\).

\[ \begin{aligned} A &= \{(1,2)\}\\ P(A) &= \frac{1}{12} \end{aligned} \]

(ii) For the sum to be \(12\), the only possible outcome is \((6,6)\).

\[ \begin{aligned} B &= \{(6,6)\}\\ P(B) &= \frac{1}{12} \end{aligned} \]

Therefore, the probability that the sum is \(3\) is \(\frac{1}{12}\), and the probability that the sum is \(12\) is also \(\frac{1}{12}\).

Q6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Solution

The city council consists of \(4\) men and \(6\) women, giving a total of \(10\) members. Since one member is selected at random, each council member has an equal chance of being chosen.

Let \(W\) denote the event that the selected member is a woman. The number of favourable outcomes is \(6\), and the total number of possible outcomes is \(10\).

\[ \begin{aligned} P(W) &= \frac{6}{10}\\ &= \frac{3}{5} \end{aligned} \]

Hence, the probability that the randomly selected council member is a woman is \(\frac{3}{5}\).

Q7. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up.
From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Solution

The coin is tossed \(4\) times, so the total number of outcomes in the sample space is \(2^4=16\). Each outcome is equally likely.

The possible results depend on the number of heads obtained. Let the number of heads be \(h\) and tails be \(4-h\). For each head there is a gain of Re \(1\), and for each tail there is a loss of Rs \(1.50\). Hence the net amount of money after four tosses is

\[ \begin{aligned} \text{Amount} &= h(1)-(4-h)(1.5) \end{aligned} \]

For \(h=4\), the outcome is \((H,H,H,H)\), giving a gain of Rs \(4\). There is only \(1\) such outcome.

\[ \begin{aligned} P(4) &= \frac{1}{16} \end{aligned} \]

For \(h=3\), the amount is \(3\times1-1\times1.5=Rs\,1.50\). There are \(^{4}C_{3}=4\) such outcomes.

\[ \begin{aligned} P(1.5) &= \frac{4}{16}=\frac{1}{4} \end{aligned} \]

For \(h=2\), the amount is \(2\times1-2\times1.5=-Rs\,1\). There are \(^{4}C_{2}=6\) such outcomes.

\[ \begin{aligned} P(-1) &= \frac{6}{16}=\frac{3}{8} \end{aligned} \]

For \(h=1\), the amount is \(1\times1-3\times1.5=-Rs\,3.50\). There are \(^{4}C_{1}=4\) such outcomes.

\[ \begin{aligned} P(-3.5) &= \frac{4}{16}=\frac{1}{4} \end{aligned} \]

For \(h=0\), the outcome is \((T,T,T,T)\), giving a loss of Rs \(6\). There is only \(1\) such outcome.

\[ \begin{aligned} P(-6) &= \frac{1}{16} \end{aligned} \]

Thus, there are five different possible amounts after four tosses: Rs \(4\), Rs \(1.50\), \(-Rs\,1\), \(-Rs\,3.50\), and \(-Rs\,6\), with respective probabilities \(\frac{1}{16},\ \frac{1}{4},\ \frac{3}{8},\ \frac{1}{4},\ \frac{1}{16}\).

Q8. Three coins are tossed once. Find the probability of getting
(i) 3 heads
(ii) 2 heads
(iii) atleast 2 heads
(iv) atmost 2 heads
(v) no head
(vi) 3 tails
(vii) exactly 2 tails
(viii) no tail
(ix) atmost 2 tails

Solution

When three fair coins are tossed once, each coin has two possible outcomes. Hence the total number of outcomes in the sample space is \(2^3=8\), and all outcomes are equally likely.

The sample space is \(\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}\).

(i) Getting \(3\) heads corresponds only to \(HHH\).

\[ \begin{aligned} P(3H) &= \frac{1}{8} \end{aligned} \]

(ii) Getting exactly \(2\) heads corresponds to \(HHT,HTH,THH\), which are \(3\) outcomes.

\[ \begin{aligned} P(2H) &= \frac{3}{8} \end{aligned} \]

(iii) At least \(2\) heads means either \(3H\) or \(2H\).

\[ \begin{aligned} P(3H)+P(2H) &= \frac{1}{8}+\frac{3}{8}\\ &= \frac{1}{2} \end{aligned} \]

(iv) At most \(2\) heads means \(0H,1H,\) or \(2H\). There is \(1\) outcome for \(0H\), \(3\) for \(1H\), and \(3\) for \(2H\).

\[ \begin{aligned} P(0H)+P(1H)+P(2H) &= \frac{1}{8}+\frac{3}{8}+\frac{3}{8}\\ &= \frac{7}{8} \end{aligned} \]

(v) No head corresponds only to \(TTT\).

\[ \begin{aligned} P(0H) &= \frac{1}{8} \end{aligned} \]

(vi) Getting \(3\) tails also corresponds to \(TTT\).

\[ \begin{aligned} P(3T) &= \frac{1}{8} \end{aligned} \]

(vii) Exactly \(2\) tails corresponds to \(HTT,THT,TTH\), which are \(3\) outcomes.

\[ \begin{aligned} P(2T) &= \frac{3}{8} \end{aligned} \]

(viii) No tail means all heads, that is \(HHH\).

\[ \begin{aligned} P(0T) &= P(3H)\\ &= \frac{1}{8} \end{aligned} \]

(ix) At most \(2\) tails means \(0T,1T,\) or \(2T\), which together account for \(1+3+3=7\) outcomes.

\[ \begin{aligned} P(0T)+P(1T)+P(2T) &= \frac{1}{8}+\frac{3}{8}+\frac{3}{8}\\ &= \frac{7}{8} \end{aligned} \]

Thus, all required probabilities follow directly from the equally likely sample space of eight outcomes.

Q9. If \(\frac{2}{11}\) is the probability of an event, what is the probability of the event ‘not A’.

Solution

Let \(A\) be the given event. The probability of the complementary event “not \(A\)” is obtained by subtracting the probability of \(A\) from \(1\), since an event and its complement together exhaust the sample space.

\[ \begin{aligned} P(A) &= \frac{2}{11}\\ P(A') &= 1-P(A)\\ &= 1-\frac{2}{11}\\ &= \frac{11-2}{11}\\ &= \frac{9}{11} \end{aligned} \]

Hence, the probability of the event “not \(A\)” is \(\frac{9}{11}\).

Q10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) a consonant

Solution

The word ASSASSINATION contains \(13\) letters in total, so the sample space has \(13\) equally likely outcomes, one for each letter position.

The vowels in the word are \(A,A,I,A,I,O\), giving a total of \(6\) vowels. Hence, the remaining \(7\) letters are consonants.

\[ \begin{aligned} P(V) &= \frac{6}{13}\\ P(C) &= \frac{7}{13} \end{aligned} \]

Therefore, the probability that the chosen letter is a vowel is \(\frac{6}{13}\), and the probability that it is a consonant is \(\frac{7}{13}\).

Q11. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

Solution

Out of the numbers \(1\) to \(20\), the lottery committee fixes exactly \(6\) distinct numbers. A player also chooses \(6\) distinct numbers at random. Since the order of selection does not matter, the total number of possible selections is \(^{20}C_{6}\). Each selection is equally likely.

There is only one favourable case, namely when the player’s chosen set exactly matches the committee’s fixed set. Hence the number of favourable outcomes is \(1\).

\[ \begin{aligned} n(E) &= 1\\ n(S) &= ^{20}C_{6}\\ P(E) &= \frac{n(E)}{n(S)}\\ &= \frac{1}{^{20}C_{6}}\\ &= \frac{1}{\dfrac{20!}{6!\,14!}}\\ &= \frac{6!\,14!}{20!}\\ &= \frac{6\times5\times4\times3\times2\times1}{20\times19\times18\times17\times16\times15}\\ &= \frac{1}{38760} \end{aligned} \]

Therefore, the probability of winning the prize in this lottery game is \(\dfrac{1}{38760}\).

Q12. Check whether the following probabilities P(A) and P(B) are consistently defined
(i) \(P(A) = 0.5, P(B) = 0.7, P(A \cap B) = 0.6\)
(ii) \(P(A) = 0.5, P(B) = 0.4, P(A \cup B) = 0.8\)

Solution

(i) Given \(P(A)=0.5\), \(P(B)=0.7\), and \(P(A\cap B)=0.6\). For any two events, the addition law gives \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), and it must satisfy \(P(A\cup B)\le 1\).

\[ \begin{aligned} P(A)+P(B)-P(A\cap B) &= 0.5+0.7-0.6\\ &= 0.6 \end{aligned} \]

Here, however, we also note that \(P(A\cap B)\) cannot exceed either \(P(A)\) or \(P(B)\). Since \(P(A\cap B)=0.6\) is greater than \(P(A)=0.5\), this violates a basic property of probabilities. Hence these values are not consistently defined.

(ii) Given \(P(A)=0.5\), \(P(B)=0.4\), and \(P(A\cup B)=0.8\). Again using the addition law,

\[ \begin{aligned} P(A\cap B) &= P(A)+P(B)-P(A\cup B)\\ &= 0.5+0.4-0.8\\ &= 0.1 \end{aligned} \]

Since \(P(A\cap B)=0.1\) is non-negative and does not exceed either \(P(A)\) or \(P(B)\), and all probabilities lie between \(0\) and \(1\), these values satisfy the required conditions.

Therefore, case (i) is not consistent, whereas case (ii) is consistent.

Q13. Fill in the blanks in following table: \[ \begin{aligned} \begin{array}{} &P(A)&P(B)&P(A\cap B)&P(A\cup B)\\ (i) &\frac{1}{3} &\frac{1}{5}&\frac{1}{15}&\ldots\\ (ii)&0.35&\ldots&0.25&0.6\\ (iii)&0.5&0.35&\ldots&0.7 \end{array} \end{aligned} \]

Solution

To fill in the missing entries, we use the standard relation between probabilities of two events, \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

For part (i), the given values are \(P(A)=\frac{1}{3}\), \(P(B)=\frac{1}{5}\), and \(P(A\cap B)=\frac{1}{15}\).

\[ \begin{aligned} P(A\cup B) &= P(A)+P(B)-P(A\cap B)\\ &= \frac{1}{3}+\frac{1}{5}-\frac{1}{15}\\ &= \frac{5+3-1}{15}\\ &= \frac{7}{15} \end{aligned} \]

For part (ii), the given values are \(P(A)=0.35\), \(P(A\cap B)=0.25\), and \(P(A\cup B)=0.6\). Using the same formula and solving for \(P(B)\),

\[ \begin{aligned} P(A\cup B) &= P(A)+P(B)-P(A\cap B)\\ 0.6 &= 0.35+P(B)-0.25\\ P(B) &= 0.6+0.25-0.35\\ &= 0.50 \end{aligned} \]

For part (iii), the given values are \(P(A)=0.5\), \(P(B)=0.35\), and \(P(A\cup B)=0.7\). Rearranging the formula to find \(P(A\cap B)\),

\[ \begin{aligned} P(A\cap B) &= P(A)+P(B)-P(A\cup B)\\ &= 0.5+0.35-0.7\\ &= 0.15 \end{aligned} \]

Hence, the missing entries are \(P(A\cup B)=\frac{7}{15}\) in (i), \(P(B)=0.50\) in (ii), and \(P(A\cap B)=0.15\) in (iii).

Q14. Given \(P(A)=\frac{3}{5}\) and \(P(B)=\frac{1}{5}\). Find P(A or B), if A and B are mutually exclusive events.

Solution

Since \(A\) and \(B\) are mutually exclusive events, they cannot occur together. Therefore, the probability of \(A\) or \(B\) occurring is simply the sum of their individual probabilities.

\[ \begin{aligned} P(A) &= \frac{3}{5}\\ P(B) &= \frac{1}{5}\\ P(A\cup B) &= P(A)+P(B)\\ &= \frac{3}{5}+\frac{1}{5}\\ &= \frac{4}{5} \end{aligned} \]

Hence, \(P(A\text{ or }B)=\frac{4}{5}\).

Q15. If E and F are events such that \(P(E)=\frac{1}{4}\) and \(P(F)=\frac{1}{2}\) and \(P(E \text{ and }F)=\frac{1}{8}\), find
(i) P(E or F),
(ii) P(not E and not F).

Solution

The given probabilities are \(P(E)=\frac{1}{4}\), \(P(F)=\frac{1}{2}\), and \(P(E\cap F)=\frac{1}{8}\). To find \(P(E\text{ or }F)\), we use the addition law of probability, \(P(E\cup F)=P(E)+P(F)-P(E\cap F)\).

\[ \begin{aligned} P(E) &= \frac{1}{4}\\ P(F) &= \frac{1}{2}\\ P(E\cap F) &= \frac{1}{8}\\ P(E\cup F) &= P(E)+P(F)-P(E\cap F)\\ &= \frac{1}{4}+\frac{1}{2}-\frac{1}{8}\\ &= \frac{2+4-1}{8}\\ &= \frac{5}{8} \end{aligned} \]

Hence,

(i) \(P(E\text{ or }F)=\frac{5}{8}\).

For part (ii), “not \(E\) and not \(F\)” is the complement of \(E\cup F\). Therefore,

\[ \begin{aligned} P(E' \cap F') &= 1-P(E\cup F)\\ &= 1-\frac{5}{8}\\ &= \frac{3}{8} \end{aligned} \]

Thus, \(P(E\text{ or }F)=\frac{5}{8}\) and \(P(\text{not }E\text{ and not }F)=\frac{3}{8}\).

Q16. Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.

Solution

It is given that \(P(\text{not }E\text{ or not }F)=0.25\). By De Morgan’s law, \(\text{not }E\text{ or not }F=(E\cap F)'\). Hence,

\[ \begin{aligned} P((E\cap F)') &= 0.25\\ P(E\cap F) &= 1-0.25\\ &= 0.75 \end{aligned} \]

Thus, \(P(E\cap F)=0.75\), which is not zero. Since mutually exclusive events must satisfy \(P(E\cap F)=0\), this condition is violated.

Therefore, \(E\) and \(F\) are not mutually exclusive.

Q17. \(A\) and \(B\) are events such that \(P(A) = 0.42,\; P(B) = 0.48\) and \(P(A and B) = 0.16\). Determine
(i) \(P\)(not \(A\)),
(ii) \(P\)(not \(B\)) and
(iii) \(P(A \text{ or } B)\)

Solution

Given that the probability of event \(A\) is \(0.42\), the probability of event \(B\) is \(0.48\), and the probability that both events occur together is \(P(A \cap B)=0.16\)

The probability of the complement of an event is obtained by subtracting the probability of the event from \(1\). Hence the required values are calculated as follows

\[ \begin{aligned} P(A) &= 0.42\\ P(B) &= 0.48\\ P(A \cap B) &= 0.16\\ P(A') &= 1 - P(A)\\ &= 1 - 0.42\\ &= 0.58\\\\ P(B') &= 1 - P(B)\\ &= 1 - 0.48\\ &= 0.52 \end{aligned} \]

To find the probability that at least one of the events \(A\) or \(B\) occurs, the addition rule of probability is used. This rule avoids double counting the common outcomes of \(A\) and \(B\)

\[ \begin{aligned} P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ &= 0.42 + 0.48 - 0.16\\ &= 0.90 - 0.16\\ &= 0.74 \end{aligned} \]

Q18. In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Solution

It is given that 40% of the students study Mathematics, 30% study Biology, and 10% of the students study both Mathematics and Biology. Converting these percentages into probabilities helps in applying the rules of probability

\[ \begin{aligned} P(M) &= 0.4\\ P(B) &= 0.3\\ P(M \cap B) &= 0.1 \end{aligned} \]

The probability that a randomly selected student studies Mathematics or Biology is found using the addition rule of probability. This rule is used to ensure that students studying both subjects are not counted twice

\[ \begin{aligned} P(M \cup B) &= P(M) + P(B) - P(M \cap B)\\ &= 0.4 + 0.3 - 0.1\\ &= 0.6 \end{aligned} \]

Q19. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both?

Solution

Let \(F\) denote the event that a student passes the first examination and \(S\) denote the event that a student passes the second examination. The given probabilities describe individual performance and the chance of passing at least one of the two examinations

\[ \begin{aligned} P(F) &= 0.8\\ P(S) &= 0.7\\ P(F \cup S) &= 0.95 \end{aligned} \]

To find the probability that a student passes both examinations, the addition rule of probability is used. This rule relates the probabilities of individual events, their union, and their intersection

\[ \begin{aligned} P(F \cap S) &= P(F) + P(S) - P(F \cup S)\\ &= 0.8 + 0.7 - 0.95\\ &= 0.55 \end{aligned} \]

Q20. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

Solution

Let \(E\) be the event that a student passes the English examination and \(H\) be the event that a student passes the Hindi examination. The probability of passing neither examination is given, which helps in finding the probability of passing at least one examination

\[ \begin{aligned} P(E) &= 0.75\\ P(E \cap H) &= 0.5\\ P(\text{neither } E \text{ nor } H) &= 0.1\\ P(E \cup H) &= 1 - 0.1\\ &= 0.9 \end{aligned} \]

Using the addition rule of probability, the probability of passing at least one of the two examinations is expressed in terms of individual probabilities and their intersection. This relation is then used to determine the probability of passing the Hindi examination

\[ \begin{aligned} P(E \cup H) &= P(E) + P(H) - P(E \cap H)\\ P(H) &= P(E \cup H) + P(E \cap H) - P(E)\\ &= 0.9 + 0.5 - 0.75\\ &= 0.65 \end{aligned} \]

Q21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.

Solution

The total number of students in the class is 60, which forms the sample space. Out of these, 30 students opted for NCC, 32 opted for NSS, and 24 students opted for both NCC and NSS. Probabilities are obtained by dividing the respective numbers by the total number of students

First, the probability that a randomly selected student opted for NCC or NSS is calculated using the addition rule of probability, which avoids counting students who opted for both activities twice

\[ \begin{aligned} P(NCC) &= \dfrac{30}{60}\\ P(NSS) &= \dfrac{32}{60}\\ P(NCC \cap NSS) &= \dfrac{24}{60}\\ P(NCC \cup NSS) &= \dfrac{30}{60} + \dfrac{32}{60} - \dfrac{24}{60}\\ &= \dfrac{38}{60}\\ &= \dfrac{19}{30} \end{aligned} \]

The probability that a student opted for neither NCC nor NSS is the complement of the event that the student opted for NCC or NSS

\[ \begin{aligned} P(\text{neither NCC nor NSS}) &= 1 - \dfrac{19}{30}\\ &= \dfrac{11}{30} \end{aligned} \]

The probability that a student opted for NSS but not NCC is found by subtracting the number of students who opted for both from those who opted for NSS only

\[ \begin{aligned} P(\text{NSS but not NCC}) &= \dfrac{32 - 24}{60}\\ &= \dfrac{8}{60}\\ &= \dfrac{2}{15} \end{aligned} \]