GSEB Class 12 Maths Solutions Chapter 13 Probability Exercise 13.5

Question 1. A dice is thrown 6 times. "If getting an odd number” is a success, what is the probability of
(iii) At most 5 successes?
Answer: A standard dice has three odd numbers. So, the chance of getting an odd number when you roll the dice is \( \frac{3}{6} = \frac{1}{2} \).
(i) Probability of getting 5 successes
\( P(5) = ^6C_5(\frac{1}{2})(\frac{1}{2})^5 \)
\( = 6.(\frac{1}{2})^6 = \frac{6}{64} = \frac{3}{32} \).
(ii) Probability of getting at least 5 successes
\( = P(\text{at least 5 successes}) \)
\( = P(5) + P(6) = ^6C_5(\frac{1}{2})^5(\frac{1}{2})^1 + ^6C_6(\frac{1}{2})^6 \)
\( = (\frac{1}{2})^6 (6 + 1) = \frac{7}{64} \).
(iii) P(at most 5 successes)
\( = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) \)
\( = [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] - P(6) \)
\( = 1 - P(6) = 1 - (\frac{1}{2})^6 = 1 - \frac{1}{64} = \frac{63}{64} \).
In simple words: First, discover the likelihood of getting an odd number when rolling a dice. Then, compute the probability of getting exactly 5 odd numbers in 6 throws. After that, figure out the probability of getting 5 or 6 odd numbers. Finally, determine the probability of getting up to 5 odd numbers by subtracting the probability of getting exactly 6 odd numbers from the total probability of 1.

Exam Tip: For binomial distribution problems, clearly identify n (number of trials), p (probability of success), and q (probability of failure) before applying the formula. Remember that "at most k" means \( P(X \le k) \), and "at least k" means \( P(X \ge k) \).

Question 2. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Answer: When a pair of dice is thrown, a doublet may be obtained in 6 ways. The total number of possible outcomes is 36. So, the probability of getting a doublet is \( \frac{6}{36} = \frac{1}{6} \). This makes \( p = \frac{1}{6} \) and \( q = \frac{5}{6} \). In 4 throws, the probability of getting two successes is:
\( = ^4C_2 p^2 q^2 = ^4C_2 (\frac{1}{6})^2 (\frac{5}{6})^2 \)
\( = \frac{4 \times 3}{1 \times 2} \times \frac{1}{36} \times \frac{25}{36} \)
\( = \frac{6 \times 25}{1296} = \frac{150}{1296} = \frac{25}{216} \).
In simple words: Find the chance of rolling two identical numbers (a doublet) with two dice. Since this is a binomial problem, figure out 'p' (success probability) and 'q' (failure probability). Then, use the binomial formula for exactly two successes in four tries.

Exam Tip: Remember that a doublet means both dice show the same number (e.g., (1,1), (2,2), etc.). Carefully count these successful outcomes and the total possible outcomes when two dice are rolled.

Question 3. There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Answer: The probability of getting one defective item is 5%, which means \( \frac{5}{100} = \frac{1}{20} \). Therefore, the probability of getting a good item is \( 1 - \frac{1}{20} = \frac{19}{20} \). A sample of 10 items includes not more than one defective item, meaning the sample contains at most one defective item. Its probability is \( P(0) + P(1) \).
\( = ^{10}C_0 (\frac{19}{20})^{10} (\frac{1}{20})^0 + ^{10}C_1 (\frac{19}{20})^9 (\frac{1}{20})^1 \)
\( = (\frac{19}{20})^{10} + 10 \times (\frac{19}{20})^9 \times \frac{1}{20} \)
\( = (\frac{19}{20})^9 (\frac{19}{20} + \frac{10}{20}) \)
\( = \frac{29}{20} (\frac{19}{20})^9 \).
In simple words: First, find the chances of having a defective item and a good item. Then, calculate the likelihood that, out of 10 chosen items, you will find either zero defective items or exactly one defective item. Add these two probabilities together to get the final answer.

Exam Tip: "Not more than one" translates to "zero or one." Ensure you correctly apply the binomial probability formula for both \( P(X=0) \) and \( P(X=1) \) and then add them together.

Question 4. Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. What is the probability that
(i) all five cards are spades?
(ii) only three cards are spades?
(iii) none is a spade?
Answer: Let one card drawn be a spade. Its probability is \( \frac{13}{52} = \frac{1}{4} \). Therefore, the probability that the card drawn is not a spade is \( 1 - \frac{1}{4} = \frac{3}{4} \).
(i) P(all cards drawn are spades)
\( = (\frac{1}{4})^5 = \frac{1}{1024} \).
(ii) P(only three cards are spades)
\( = ^5C_3 (\frac{3}{4})^2 (\frac{1}{4})^3 \)
\( = \frac{5 \times 4}{2 \times 1} \times \frac{9}{16} \times \frac{1}{64} \)
\( = 10 \times \frac{9}{1024} = \frac{90}{1024} = \frac{45}{512} \).
(iii) P(none is a spade)
\( = (\frac{3}{4})^5 = \frac{243}{1024} \).
In simple words: Calculate the chance of picking a spade and not picking a spade. Then, for each part of the question, use the binomial probability formula. For "all five spades," it means five successes. For "only three spades," it means three successes and two failures. For "none is a spade," it means five failures.

Exam Tip: Since cards are drawn "with replacement," the probability of drawing a spade (or not a spade) remains constant for each draw. This indicates a binomial distribution problem.

Question 5. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs,
(iii) more than one
(iv) at least one will fuse after 150 days of use.
Answer: The probability that a bulb gets fused after 150 days of its use is 0.05. Therefore, the probability that the bulb will not fuse after 150 days of its use is \( 1 - 0.05 = 0.95 \).
(i) Probability that no bulb will fuse after 150 days of its use
\( = P(\text{none}) = (0.95)^5 \)
\( = 0.7738 \approx 0.77 \).
(ii) P(not more than one)
\( = P(0) + P(1) \)
\( = (0.95)^5 + ^5C_1 \times (0.95)^4 \times (0.05) \)
\( = (0.95)^4 [0.95 + 5 \times 0.05] \)
\( = (0.95)^4 (0.95 + 0.25) \)
\( = (0.95)^4 \times 1.2 \).
(iii) P(more than one)
\( = P(2) + P(3) + P(4) + P(5) \)
\( = [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] - [P(0) + P(1)] \)
\( = 1 - [P(0) + P(1)] \)
\( = 1 - (0.95)^4 \times 1.2 \) [from part (ii)]
(iv) P(at least one)
\( = P(1) + P(2) + P(3) + P(4) + P(5) \)
\( = 1 - P(0) \)
\( = 1 - (0.95)^5 \) [from part (i)]
In simple words: First, find the probability of a bulb failing and not failing. For part (iii), "more than one" means more than one bulb fails, so subtract the probabilities of zero or one failure from the total. For part (iv), "at least one" means one or more bulbs fail, so subtract the probability of zero failures from the total.

Exam Tip: Remember to use the complement rule for "more than one" and "at least one" scenarios, as it often simplifies calculations significantly (i.e., \( P(X > k) = 1 - P(X \le k) \) and \( P(X \ge k) = 1 - P(X < k) \)).

Question 6. A bag consists of 10 bulbs each marked with one of the digits 0 to 9. If four bulbs are drawn successively with replacement from the bag, what is the probability that none is marked with digit 0?
Answer: A bag contains 10 bulbs marked from 0 to 9. The probability of getting a bulb marked with 0 is \( \frac{1}{10} = 0.1 \). Therefore, the probability that the bulb marked 0 is not obtained is \( 1 - 0.1 = 0.9 \). Now, 4 bulbs are drawn. The probability that none is marked with 0 is:
\( = (0.9)^4 = (\frac{9}{10})^4 \).
In simple words: First, calculate the chance of picking a bulb that is *not* marked with 0. Since you draw four bulbs one after another, and you replace them each time, multiply this probability by itself four times. This gives you the likelihood that none of the four bulbs will have a 0 on them.

Exam Tip: When items are drawn "with replacement," the probability of an event remains constant for each trial, making it a Bernoulli trial or binomial distribution scenario.

Question 7. In an examination, 20 true-false type questions are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls head, he answer "true', if it falls tail, he answers 'false'. Find the probability that he answers at least 12 questions correctly.
Answer: The probability that a student answers a question true is \( \frac{1}{2} \). This means when a coin is thrown, the probability that a head is obtained is \( \frac{1}{2} \). The probability that his answer is false is \( 1 - \frac{1}{2} = \frac{1}{2} \). The probability that he answers at least 12 questions correctly is:
\( = P(12) + P(13) + P(14) + \dots + P(20) \)
\( = ^{20}C_{12}(\frac{1}{2})^{12}(\frac{1}{2})^8 + ^{20}C_{13}(\frac{1}{2})^{13}(\frac{1}{2})^7 + \dots + ^{20}C_{20}(\frac{1}{2})^{20}(\frac{1}{2})^0 \)
\( = (\frac{1}{2})^{20} [^{20}C_{12} + ^{20}C_{13} + \dots + ^{20}C_{20}] \).
In simple words: Since a fair coin is tossed, the chance of answering any question correctly is 1/2. To find the probability of answering at least 12 questions correctly out of 20, you need to sum the probabilities of answering exactly 12, 13, ..., all the way up to 20 questions correctly.

Exam Tip: For true-false questions answered by a coin toss, the probability of success (answering correctly) is always 1/2. Remember that \( (\frac{1}{2})^n \) can be factored out when summing probabilities in such cases.

Question 8. Suppose X has a binomial distribution B (6, \( \frac{1}{2} \)), show that X = 3 is mot likely outcome.
Answer: For a binomial distribution \( B(n, p) \), the most likely outcome is \( X = np \) or the integer closest to it. Here, \( n = 6 \) and \( p = \frac{1}{2} \).
\( np = 6 \times \frac{1}{2} = 3 \).
Thus, \( X = 3 \) is the most likely outcome.
Alternatively, we calculate the probabilities for \( X=0, 1, 2, 3, 4, 5, 6 \):
\( P(X=k) = ^nC_k p^k (1-p)^{n-k} \)
When \( n=6, p=\frac{1}{2} \), then \( 1-p = \frac{1}{2} \).
\( P(X=k) = ^6C_k (\frac{1}{2})^k (\frac{1}{2})^{6-k} = ^6C_k (\frac{1}{2})^6 \).
To find the maximum probability, we need to find the maximum value of \( ^6C_k \).
The binomial coefficients \( ^6C_k \) are: \( ^6C_0 = 1, ^6C_1 = 6, ^6C_2 = 15, ^6C_3 = 20, ^6C_4 = 15, ^6C_5 = 6, ^6C_6 = 1 \).
The maximum value among these is \( ^6C_3 = 20 \).
Therefore, \( P(X=3) = ^6C_3 (\frac{1}{2})^6 = 20 \times \frac{1}{64} = \frac{20}{64} = \frac{5}{16} \), which is the maximum probability.
This shows that \( X=3 \) is the most likely outcome.
In simple words: In a binomial distribution, the most probable outcome (the mode) occurs when the number of successes is equal to, or closest to, the product of the number of trials and the probability of success (\( np \)). Here, \( np \) is 3, so 3 successes are the most likely. We can confirm this by checking all the probabilities, and the one for 3 successes is indeed the highest.

Exam Tip: For a binomial distribution \( B(n, p) \), when \( p = \frac{1}{2} \), the distribution is symmetric, and the most likely outcome is exactly \( n/2 \) if n is even, or the two integers closest to \( n/2 \) if n is odd.

Question 9. On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers, just by guessing?
Answer: One option out of three is correct. The probability that the answer is correct, \( p \), is \( \frac{1}{3} \). The probability that the answer is incorrect is \( 1 - \frac{1}{3} = \frac{2}{3} = q \). The probability that four or more answers are correct is:
\( = P(4) + P(5) \)
\( = ^5C_4 p^4 q^1 + ^5C_5 p^5 q^0 \)
\( = ^5C_4 (\frac{1}{3})^4 (\frac{2}{3})^1 + ^5C_5 (\frac{1}{3})^5 (\frac{2}{3})^0 \)
\( = 5 \times \frac{1}{81} \times \frac{2}{3} + 1 \times \frac{1}{243} \times 1 \)
\( = \frac{10}{243} + \frac{1}{243} \)
\( = \frac{11}{243} \).
In simple words: Determine the chance of guessing a question right (which is 1 out of 3 choices). Then, calculate the chance of getting exactly four questions right, and add that to the chance of getting all five questions right. This sum will give you the total likelihood of getting four or more correct answers.

Exam Tip: "Four or more" means \( P(X \ge 4) \), which simplifies to \( P(X=4) + P(X=5) \) for five questions. Make sure to identify 'n', 'p', and 'q' correctly for the binomial formula.

Question 10. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \( \frac{1}{100} \). What is the probability that he will win a prize
(a) at least once?
(b) exactly once?
(c) at least twice?
Answer: The probability that the person wins the prize is \( p = \frac{1}{100} \). The probability of losing is \( q = 1 - \frac{1}{100} = \frac{99}{100} \). The number of lotteries (trials) is \( n=50 \).
(a) Probability of winning at least once
\( = 1 - P(\text{losing in all 50 lotteries}) \)
\( = 1 - (\frac{99}{100})^{50} \)
\( = 1 - (0.99)^{50} \).
(b) Probability that he wins exactly once
\( = P(X=1) = ^{50}C_1 (\frac{1}{100})^1 (\frac{99}{100})^{49} \)
\( = 50 \times \frac{1}{100} \times (\frac{99}{100})^{49} \)
\( = \frac{1}{2} (\frac{99}{100})^{49} \).
(c) Probability that he wins at least twice
\( = P(X \ge 2) = 1 - [P(0) + P(1)] \)
\( = 1 - [^{50}C_0 (\frac{99}{100})^{50} (\frac{1}{100})^0 + ^{50}C_1 (\frac{99}{100})^{49} (\frac{1}{100})^1] \)
\( = 1 - [(\frac{99}{100})^{50} + 50 \times \frac{1}{100} \times (\frac{99}{100})^{49}] \)
\( = 1 - [(\frac{99}{100})^{50} + \frac{1}{2} (\frac{99}{100})^{49}] \)
\( = 1 - (\frac{99}{100})^{49} (\frac{99}{100} + \frac{1}{2}) \)
\( = 1 - (\frac{99}{100})^{49} (\frac{99+50}{100}) \)
\( = 1 - (\frac{99}{100})^{49} (\frac{149}{100}) \).
In simple words: First, determine the chance of winning and losing a single lottery. Then, for part (a), subtract the probability of losing all 50 lotteries from 1. For part (b), use the binomial formula to find the exact probability of winning exactly once. For part (c), subtract the probabilities of winning zero times and winning exactly once from 1.

Exam Tip: Remember that "at least once" is \( 1 - P(\text{none}) \) and "at least twice" is \( 1 - [P(0) + P(1)] \). These complement rules often make calculations simpler for binomial distributions.

Question 11. Find the probability of getting 5 exactly twice in 7 throws of a dice.
Answer: When a dice is thrown, the probability of getting a 5 is \( p = \frac{1}{6} \). Therefore, the probability of not getting a 5 is \( q = 1 - \frac{1}{6} = \frac{5}{6} \). We have 7 throws, so \( n=7 \). The probability of getting 5 exactly twice in 7 throws is:
\( = P(X=2) = ^7C_2 (\frac{1}{6})^2 (\frac{5}{6})^5 \)
\( = \frac{7 \times 6}{1 \times 2} \times \frac{1}{36} \times \frac{3125}{7776} \)
\( = 21 \times \frac{1}{36} \times \frac{3125}{7776} \)
\( = \frac{7}{12} \times \frac{3125}{7776} \)
\( = \frac{21875}{93312} \).
In simple words: First, calculate the chance of rolling a 5 and the chance of not rolling a 5. Then, use the binomial probability formula for getting exactly two successes (rolling a 5) in seven attempts.

Exam Tip: For "exactly k successes," directly apply the binomial probability formula \( P(X=k) = ^nC_k p^k q^{n-k} \). Ensure you correctly identify 'n', 'k', 'p', and 'q'.

Question 12. Find the probability of throwing at most 2 sixes in 6 throws of a single dice.
Answer: When a dice is thrown, the probability of getting a six is \( p = \frac{1}{6} \). Therefore, the probability of not getting a six is \( q = 1 - \frac{1}{6} = \frac{5}{6} \). We have 6 throws, so \( n=6 \). The probability of throwing at most 2 sixes in 6 throws of a single dice is:
\( = P(0) + P(1) + P(2) \)
\( = ^6C_0 (\frac{1}{6})^0 (\frac{5}{6})^6 + ^6C_1 (\frac{1}{6})^1 (\frac{5}{6})^5 + ^6C_2 (\frac{1}{6})^2 (\frac{5}{6})^4 \)
\( = 1 \times 1 \times (\frac{5}{6})^6 + 6 \times \frac{1}{6} \times (\frac{5}{6})^5 + \frac{6 \times 5}{2 \times 1} \times \frac{1}{36} \times (\frac{5}{6})^4 \)
\( = (\frac{5}{6})^6 + (\frac{5}{6})^5 + 15 \times \frac{1}{36} \times (\frac{5}{6})^4 \)
\( = (\frac{5}{6})^4 [(\frac{5}{6})^2 + (\frac{5}{6})^1 + \frac{15}{36}] \)
\( = (\frac{5}{6})^4 [\frac{25}{36} + \frac{5}{6} + \frac{15}{36}] \)
\( = (\frac{5}{6})^4 [\frac{25 + 30 + 15}{36}] \)
\( = (\frac{5}{6})^4 \frac{70}{36} \)
\( = \frac{625}{1296} \times \frac{70}{36} = \frac{625 \times 35}{648 \times 18} = \frac{21875}{11664} \).
In simple words: First, calculate the probability of rolling a six and not rolling a six. Then, since "at most 2 sixes" means zero, one, or two sixes, you must calculate the probabilities for each of these three cases separately and then add them together to get the final result.

Exam Tip: "At most 2" means \( P(X \le 2) = P(X=0) + P(X=1) + P(X=2) \). Be careful with the arithmetic when summing multiple binomial probabilities.

Question 13. It is known that 10% of a certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective.
Answer: The probability that the article selected is defective is \( p = 10\% = \frac{1}{10} \). Therefore, the probability that the article selected is good is \( q = 1 - \frac{1}{10} = \frac{9}{10} \). In a sample of 12 articles, we want to find the probability that 9 are defective:
\( = ^{12}C_9 (\frac{1}{10})^9 (\frac{9}{10})^3 \)
\( = ^{12}C_3 (\frac{1}{10})^9 (\frac{9}{10})^3 \)
\( = \frac{12 \times 11 \times 10}{1 \times 2 \times 3} \times \frac{1}{10^9} \times \frac{9^3}{10^3} \)
\( = 220 \times \frac{729}{10^{12}} \)
\( = \frac{160380}{10^{12}} = \frac{16038}{10^{11}} \).
In simple words: First, determine the probability of an item being defective and the probability of it being good. Then, use the binomial probability formula to calculate the exact chance of finding 9 defective items in a group of 12 items.

Exam Tip: Remember the property \( ^nC_k = ^nC_{n-k} \) to simplify calculations (e.g., \( ^{12}C_9 = ^{12}C_3 \)). This can make expanding the combination term easier.

Question 14. In a sample of 5 bulbs from a box containing 100 bulbs (10 of which are defective), the probability that none is defective is
(A) \( 10^{-1} \)
(B) \( (\frac{1}{2})^5 \)
(C) \( (\frac{9}{10})^5 \)
(D) \( (\frac{9}{10}) \)
Answer: (C) \( (\frac{9}{10})^5 \)
In simple words: First, calculate the probability that a single bulb chosen randomly is not defective. Since you are choosing 5 bulbs and want none to be defective, multiply this probability by itself 5 times.

Exam Tip: The context describes a situation where a smaller sample (5 bulbs) is drawn from a larger population (100 bulbs) with replacement (implied by typical binomial distribution problems unless stated otherwise). The probability of an item being non-defective is constant for each pick.

Question 15. The probability that a student is not a swimmer is \( \frac{1}{5} \). Then, the probability that out of five students, four are swimmers is
(A) \( ^5C_4(\frac{4}{5})^4(\frac{1}{5})^1 \)
(B) \( (\frac{4}{5})^4(\frac{1}{5})^1 \)
(C) \( ^5C_1(\frac{4}{5})^1(\frac{1}{5})^4 \)
(D) None of these
Answer: (A) \( ^5C_4(\frac{4}{5})^4(\frac{1}{5})^1 \)
In simple words: If the chance of not being a swimmer is 1/5, then the chance of being a swimmer is 4/5. To find the probability that exactly four out of five students are swimmers, use the binomial probability formula: (number of ways to choose 4 from 5) multiplied by (probability of being a swimmer)^4 multiplied by (probability of not being a swimmer)^1.

Exam Tip: Clearly define 'p' as the probability of "success" (being a swimmer in this case) and 'q' as the probability of "failure" (not being a swimmer). Apply the binomial formula \( ^nC_k p^k q^{n-k} \) directly.