Maharashtra Board Class 12 Maths Part 2 Chapter 8 Probability Distributions 8.3 Solutions

Std 12 Maths 2 Exercise 8.3 Solutions Commerce Maths

Question 1. A die is thrown 4 times. If 'getting an odd number' is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Answer:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
\( \therefore n = 4 \)
\( \therefore p = \frac{3}{6} = \frac{1}{2} \)
\( \therefore q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \therefore X \sim B(3, \frac{1}{2}) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
(i) P(Two Successes)
\[ P(X = 2) = ^4C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{4-2} \] \[ = 6 \times \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 \] \[ = 6 \times \left(\frac{1}{2}\right)^4 \] \[ = 6 \times \frac{1}{16} \] \[ = \frac{3}{8} \]
(ii) P(Atleast 3 Successes)
\[ P(X \ge 3) = p(3) + p(4) \] \[ = ^4C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{4-3} + ^4C_4 \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{4-4} \] \[ = 4 \times \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^1 + 1 \times \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^0 \] \[ = 4 \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 \] \[ = \left(\frac{1}{2}\right)^4 (4+1) \] \[ = \frac{5}{16} \]
(iii) P(Atmost 2 Successes)
\[ P(X \le 2) = p(0) + p(1) + p(2) \] \[ = ^4C_0 \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{4-0} + ^4C_1 \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{4-1} + ^4C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{4-2} \] \[ = 1 \times 1 \times \left(\frac{1}{2}\right)^4 + 4 \times \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 + 6 \times \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 \] \[ = \left(\frac{1}{2}\right)^4 + 4 \left(\frac{1}{2}\right)^4 + 6 \left(\frac{1}{2}\right)^4 \] \[ = \left(\frac{1}{2}\right)^4 (1+4+6) \] \[ = \frac{11}{16} \]
In simple words: This problem uses the binomial probability distribution to calculate the chances of getting a specific number of odd outcomes when a die is rolled 4 times, applying the formula for exactly, at least, and at most two successes.

🎯 Exam Tip: Remember to correctly identify 'n' (number of trials) and 'p' (probability of success) in binomial distribution problems. Clearly state the condition for success and failure, and apply the correct formula for 'at least' or 'at most' scenarios by summing probabilities of relevant outcomes.

Question 2. A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Answer:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
\( \therefore p = \frac{6}{36} = \frac{1}{6} \)
\( \therefore q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \)
\( X \sim B(3, \frac{1}{6}) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
P(getting two successes)
\[ = P(X=2) \] \[ = ^3C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{3-2} \] \[ = 3 \times \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^1 \] \[ = 3 \times \frac{1}{36} \times \frac{5}{6} \] \[ = \frac{15}{216} \] \[ = \frac{5}{72} \]
In simple words: This problem calculates the probability of obtaining exactly two doublets in three throws of a pair of dice, using the binomial probability formula where getting a doublet is considered a success.

🎯 Exam Tip: For problems involving dice, always list the total possible outcomes (sample space) and the favorable outcomes for 'success' to correctly determine 'p'. Pay attention to the number of trials 'n' and the desired number of successes 'x'.

Question 3. There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Answer:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
\( \therefore p = 0.1 \)
\( \therefore q = 1 - p = 1 - 0.1 = 0.9 \)
\( X \sim B(4, 0.1) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
P(Not include more than 1 defective)
\[ P(X \le 1) = p(0) + p(1) \] \[ = ^4C_0 (0.1)^0 (0.9)^4 + ^4C_1 (0.1)^1 (0.9)^{4-1} \] \[ = 1 \times 1 \times (0.9)^4 + 4 \times 0.1 \times (0.9)^3 \] \[ = (0.9)^3 [0.9 + 0.4] \] \[ = (0.9)^3 \times 1.3 \] \[ = 0.977 \]
In simple words: This solution finds the probability of having at most one defective item in a sample of four, given that 10% of items are defective, by summing the probabilities of zero and one defective item.

🎯 Exam Tip: When dealing with percentages, convert them to decimal probabilities for 'p'. For "not more than" (i.e., at most) scenarios, remember to sum the probabilities from zero up to the specified limit.

Question 4. Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Answer:
X: No. of spade cards
Number of cards drawn
\( \therefore n = 5 \)
p: Probability of getting spade card
\( \therefore p = \frac{13}{52} = \frac{1}{4} \)
\( \therefore q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \)
\( X \sim B(5, \frac{1}{4}) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
(i) P(All five cards are spades)
\[ P(X = 5) = ^5C_5 \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^{5-5} \] \[ = 1 \times \left(\frac{1}{4}\right)^5 \times \left(\frac{3}{4}\right)^0 \] \[ = 1 \times \frac{1}{1024} \times 1 \] \[ = \frac{1}{1024} \]
(ii) P(Only 3 cards are spades)
\[ P(X = 3) = ^5C_3 \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^{5-3} \] \[ = 10 \times \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^2 \] \[ = 10 \times \frac{1}{64} \times \frac{9}{16} \] \[ = \frac{90}{1024} \] \[ = \frac{45}{512} \]
(iii) P(None is a spade)
\[ P(X = 0) = ^5C_0 \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^{5-0} \] \[ = 1 \times 1 \times \left(\frac{3}{4}\right)^5 \] \[ = \frac{243}{1024} \]
In simple words: This problem calculates the probabilities for different numbers of spades drawn in five trials with replacement, applying the binomial distribution to find the chances of all five being spades, exactly three being spades, or none being spades.

🎯 Exam Tip: In card problems, identify the total cards and the specific cards that constitute a 'success' to find 'p'. The term "with replacement" is crucial as it ensures 'p' remains constant for each trial, a prerequisite for binomial distribution.

Question 5. The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Answer:
X : No. of bulbs fuse after 200 days of use
p: Probability of getting fuse bulbs
No. of bulbs in a sample
\( \therefore n = 5 \)
\( \therefore p = 0.2 \)
\( \therefore q = 1 - p = 1 - 0.2 = 0.8 \)
\( \therefore X \sim B(5, 0.2) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
(i) P(X = 0)
\[ = ^5C_0 (0.2)^0 (0.8)^{5-0} \] \[ = 1 \times 1 \times (0.8)^5 \] \[ = (0.8)^5 \]
(ii) P(X ≤ 1)
\[ = p(0) + p(1) \] \[ = ^5C_0 (0.2)^0 (0.8)^{5-0} + ^5C_1 (0.2)^1 (0.8)^{5-1} \] \[ = 1 \times 1 \times (0.8)^5 + 5 \times 0.2 \times (0.8)^4 \] \[ = (0.8)^4 [0.8 + 1] \] \[ = 1.8 \times (0.8)^4 \]
(iii) P(X > 1)
\[ = 1 - [p(0) + p(1)] \] \[ = 1 - 1.8 \times (0.8)^4 \]
(iv) P(X ≥ 1)
\[ = 1 - p(0) \] \[ = 1 - (0.8)^5 \]
In simple words: This problem uses the binomial distribution to calculate the probability of bulbs fusing within 200 days for different conditions (zero, at most one, more than one, or at least one), given a 0.2 probability of a single bulb fusing.

🎯 Exam Tip: Pay close attention to the inequality signs in probability questions (e.g., X=0, X≤1, X>1, X≥1) and correctly apply the complementary probability rule (P(A) = 1 - P(A')) when it simplifies calculations, especially for 'greater than' or 'at least' scenarios.

Question 6. 10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Answer:
X: No. of balls drawn marked with the digit 0
n: No. of balls drawn
\( \therefore n = 4 \)
p: Probability of balls marked with 0.
\( \therefore p = \frac{1}{10} \)
\( \therefore q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10} \)
\( p(x) = ^nC_x p^x q^{n-x} \)
P(None of the ball is marked with digit 0)
\[ P(X = 0) = ^4C_0 \left(\frac{1}{10}\right)^0 \left(\frac{9}{10}\right)^{4-0} \] \[ = 1 \times 1 \times \left(\frac{9}{10}\right)^4 \] \[ = \left(\frac{9}{10}\right)^4 \]
In simple words: This solution finds the probability that none of four balls drawn with replacement are marked '0', given that there are 10 balls marked 0-9, using the binomial probability formula for zero successes.

🎯 Exam Tip: Always clearly define 'p' and 'q' based on the success condition. When asked for 'none', it means X=0, making the calculation straightforward with ^nC_0 = 1.

Question 7. In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Answer:
n: No. of Questions
\( \therefore n = 5 \)
X: No. of correct answers by guessing
p: Probability of getting correct answers
\( \therefore p = \frac{1}{3} \)
\( \therefore q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \)
\( X \sim B(5, \frac{1}{3}) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
P(Four or more correct answers)
\[ P(X \ge 4) = p(4) + p(5) \] \[ = ^5C_4 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{5-4} + ^5C_5 \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{5-5} \] \[ = 5 \times \frac{1}{3^4} \times \frac{2}{3^1} + 1 \times \frac{1}{3^5} \times \frac{2^0}{3^0} \] \[ = \frac{10}{3^5} + \frac{1}{3^5} \] \[ = \frac{11}{243} \]
In simple words: This problem calculates the probability of a student guessing four or five correct answers on a five-question multiple-choice test with three options per question, using the binomial distribution.

🎯 Exam Tip: For multiple-choice questions, the probability of guessing a correct answer ('p') is 1 divided by the number of options. For "four or more," remember to sum the probabilities of exactly four and exactly five correct answers.

Question 8. Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Answer:
X: No. of sixes in 6 throws
n: No. of times dice thrown
\( \therefore n = 6 \)
p: Probability of getting six
\( \therefore p = \frac{1}{6} \)
\( \therefore q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \)
\( \therefore X \sim B(6, \frac{1}{6}) \)
\( \therefore p(x) = ^nC_x p^x q^{n-x} \)
P(At most 2 sixes)
\[ P(X \le 2) = p(0) + p(1) + p(2) \] \[ = ^6C_0 \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^{6-0} + ^6C_1 \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{6-1} + ^6C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{6-2} \] \[ = 1 \times 1 \times \left(\frac{5}{6}\right)^6 + 6 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^5 + 15 \times \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^4 \] \[ = \left(\frac{5}{6}\right)^6 + \left(\frac{5}{6}\right)^5 + 15 \times \frac{1}{36} \times \left(\frac{5}{6}\right)^4 \] \[ = \left(\frac{5}{6}\right)^6 + \left(\frac{5}{6}\right)^5 + \frac{15}{36} \left(\frac{5}{6}\right)^4 \] \[ = \left(\frac{5}{6}\right)^4 \left[ \left(\frac{5}{6}\right)^2 + \left(\frac{5}{6}\right)^1 + \frac{15}{36} \right] \] \[ = \left(\frac{5}{6}\right)^4 \left[ \frac{25}{36} + \frac{30}{36} + \frac{15}{36} \right] \] \[ = \left(\frac{5}{6}\right)^4 \left[ \frac{25+30+15}{36} \right] \] \[ = \left(\frac{5}{6}\right)^4 \left[ \frac{70}{36} \right] \] \[ = \left(\frac{5}{6}\right)^4 \frac{35}{18} \]
In simple words: This solution determines the probability of rolling a six at most twice in six throws of a single die by summing the probabilities of getting zero, one, or two sixes using the binomial distribution.

🎯 Exam Tip: Be careful with calculations involving powers and fractions, especially when common factors can be taken out to simplify. Ensure all terms for "at most" are included in the sum.

Question 9. Given that X ~ B(n, p), (i) if n = 10 and p = 0.4, find E(X) and Var(X). (ii) if p = 0.6 and E(X) = 6, find n and Var(X). (iii) if n = 25, E(X) = 10, find p and Var(X). (iv) if n = 10, E(X) = 8, find Var(X).
Answer:
\( \therefore X \sim B (n, p), E(X) = np, V(X) = npq, q = 1 - p \)
(i) E(X) = np = 10 × 0.4 = 4
\( \therefore q = 1 - p = 1 - 0.4 = 0.6 \)
V(X) = npq = 10 × 0.4 × 0.6 = 2.4
(ii) \( \therefore p = 0.6 \)
\( \therefore q = 1 - p = 1 - 0.6 = 0.4 \)
E(X) = np
\( \therefore 6 = n \times 0.6 \)
\( \therefore n = 10 \)
\( \therefore V(X) = npq = 10 \times 0.6 \times 0.4 = 2.4 \)
(iii) E(X) = np
\( \therefore 10 = 25 \times p \)
\( \therefore p = 0.4 \)
\( \therefore q = 1, p = 1 - 0.4 = 0.6 \)
\( \therefore S.D.(X) = \sqrt{V(X)} \)
\( = \sqrt{npq} \)
\( = \sqrt{25 \times 0.4 \times 0.6} \)
\( = \sqrt{6} \)
\( = 2.4494 \)
(iv) \( \therefore E(X) = np \)
\( \therefore 8 = 10p \)
\( \therefore p = 0.8 \)
\( \therefore q = 1 - p = 1 - 0.8 = 0.2 \)
\( \therefore V(X) = npq = 10 \times 0.8 \times 0.2 = 1.6 \)
In simple words: This question demonstrates how to calculate the expected value (E(X)), variance (V(X)), and standard deviation (S.D.(X)) of a binomial distribution using the formulas E(X)=np and V(X)=npq, given different values of n, p, or E(X).

🎯 Exam Tip: Memorize the formulas for E(X)=np and V(X)=npq for binomial distributions. These are fundamental for solving problems related to the mean and spread of binomial outcomes. Remember that S.D.(X) is the square root of V(X).

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