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Probability - Random variable and its probability distribution

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A Random Variable XX is a real-valued function whose domain is the sample space SS of a random experiment. Visually, imagine a mapping where every outcome in the sample space (like 'Heads' or 'Tails') is connected by an arrow to a specific number on the real number line.

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The Probability Distribution of a random variable XX consists of the values of XX along with their corresponding probabilities P(X=xi)P(X = x_i). This can be visualized as a table with two rows or a bar graph where the heights of the bars represent the probability pip_i of each outcome xix_i.

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For a discrete probability distribution to be valid, two conditions must be met: each probability P(xi)P(x_i) must be non-negative (0≀P(xi)≀10 \le P(x_i) \le 1) and the sum of all probabilities must equal exactly 1 (βˆ‘P(xi)=1\sum P(x_i) = 1). If viewed as a pie chart, the slices representing each outcome must perfectly complete a full circle.

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The Mean or Expected Value, denoted as E(X)E(X) or ΞΌ\mu, is the weighted average of the values of XX. In a physical sense, if you were to place weights proportional to the probabilities on a lever at positions xix_i, the mean ΞΌ\mu would be the 'center of mass' or the balance point on the horizontal axis.

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Variance, denoted as Var(X)Var(X) or Οƒ2\sigma^2, measures the dispersion or spread of the random variable's values around the mean. A distribution with a tall, narrow histogram has low variance, while a short, wide histogram indicates that the values are more spread out from the mean.

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Standard Deviation Οƒ\sigma is the positive square root of the variance. It provides a measure of spread in the same units as the random variable itself, helping to visualize the 'width' of the distribution relative to the center.

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The Binomial Distribution is a specific discrete distribution where an experiment consists of nn independent trials, each having only two possible outcomes: 'Success' (probability pp) or 'Failure' (probability q=1βˆ’pq = 1-p). The graph of a binomial distribution becomes increasingly symmetrical and bell-shaped as the number of trials nn increases.

πŸ“Formulae

Total Probability Condition: βˆ‘i=1nP(xi)=1\sum_{i=1}^{n} P(x_i) = 1

Mean (Expected Value): E(X)=ΞΌ=βˆ‘i=1nxiP(xi)E(X) = \mu = \sum_{i=1}^{n} x_i P(x_i)

Variance: Var(X)=Οƒ2=E(X2)βˆ’[E(X)]2Var(X) = \sigma^2 = E(X^2) - [E(X)]^2

Expectation of X2X^2: E(X2)=βˆ‘i=1nxi2P(xi)E(X^2) = \sum_{i=1}^{n} x_i^2 P(x_i)

Standard Deviation: Οƒ=Var(X)\sigma = \sqrt{Var(X)}

Binomial Distribution: P(X=r)=(nr)prqnβˆ’rP(X = r) = \binom{n}{r} p^r q^{n-r}, where q=1βˆ’pq = 1-p

πŸ’‘Examples

Problem 1:

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let XX be the number of aces obtained. Find the probability distribution and the mean of XX.

Solution:

Let XX denote the number of aces. Possible values for XX are 0, 1, and 2. Probability of drawing an ace p=452=113p = \frac{4}{52} = \frac{1}{13}. Probability of not drawing an ace q=1βˆ’113=1213q = 1 - \frac{1}{13} = \frac{12}{13}. \n1. P(X=0)=P(Non-AceΒ andΒ Non-Ace)=1213Γ—1213=144169P(X=0) = P(\text{Non-Ace and Non-Ace}) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169} \n2. P(X=1)=P(AceΒ andΒ Non-Ace)+P(Non-AceΒ andΒ Ace)=(113Γ—1213)+(1213Γ—113)=24169P(X=1) = P(\text{Ace and Non-Ace}) + P(\text{Non-Ace and Ace}) = (\frac{1}{13} \times \frac{12}{13}) + (\frac{12}{13} \times \frac{1}{13}) = \frac{24}{169} \n3. P(X=2)=P(AceΒ andΒ Ace)=113Γ—113=1169P(X=2) = P(\text{Ace and Ace}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169} \nMean E(X)=βˆ‘xiP(xi)=0(144169)+1(24169)+2(1169)=24+2169=26169=213E(X) = \sum x_i P(x_i) = 0(\frac{144}{169}) + 1(\frac{24}{169}) + 2(\frac{1}{169}) = \frac{24 + 2}{169} = \frac{26}{169} = \frac{2}{13}.

Explanation:

We first identify the random variable XX and its possible values. Since the cards are replaced, the trials are independent. We calculate the probability for each case (0, 1, or 2 aces) and then apply the mean formula by summing the products of the values and their probabilities.

Problem 2:

Find the variance of the number obtained on a throw of a fair die.

Solution:

Sample space S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}. Let XX be the number on the die. \nP(X=1)=P(X=2)=P(X=3)=P(X=4)=P(X=5)=P(X=6)=16P(X=1) = P(X=2) = P(X=3) = P(X=4) = P(X=5) = P(X=6) = \frac{1}{6}. \nE(X)=βˆ‘xiP(xi)=16(1+2+3+4+5+6)=216=3.5E(X) = \sum x_i P(x_i) = \frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6} = 3.5. \nE(X2)=βˆ‘xi2P(xi)=16(12+22+32+42+52+62)=1+4+9+16+25+366=916E(X^2) = \sum x_i^2 P(x_i) = \frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2) = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}. \nVar(X)=E(X2)βˆ’[E(X)]2=916βˆ’(72)2=916βˆ’494=182βˆ’14712=3512β‰ˆ2.917Var(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.917.

Explanation:

This problem uses the variance formula E(X2)βˆ’[E(X)]2E(X^2) - [E(X)]^2. First, calculate the expected value (mean), then find the expected value of the squares of the outcomes, and finally subtract the square of the mean from the expected square.