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A coin is tossed twice. Let X be a random variable defined as number of heads minus number of tails. Obtain the probability distribution of X and also find its mean.
REMEMBER KNOWLEDGE 3 Marks
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Calculation / Logic
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MEDIUM
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Q: A coin is tossed twice. Let X be a random variable defined as number of heads minus number of tails. Obtain the probability distribution of X and also find its mean.

Question Analysis & Solution

Detailed Solution

Step 1: Define the Sample Space

When a coin is tossed twice, the sample space S is given by: S = {HH, HT, TH, TT}

Step 2: Define the Random Variable X

Let X be the random variable representing the number of heads minus the number of tails.

Step 3: Determine the Values of X for each outcome

For each outcome in the sample space, we calculate the value of X: - HH: 2 heads, 0 tails. X = 2 - 0 = 2 - HT: 1 head, 1 tail. X = 1 - 1 = 0 - TH: 1 head, 1 tail. X = 1 - 1 = 0 - TT: 0 heads, 2 tails. X = 0 - 2 = -2

Step 4: Determine the Probability Distribution of X

The possible values of X are -2, 0, and 2. We calculate the probabilities for each value: - P(X = -2) = P(TT) = 1/4 - P(X = 0) = P(HT) + P(TH) = 1/4 + 1/4 = 2/4 = 1/2 - P(X = 2) = P(HH) = 1/4 The probability distribution of X is: X | -2 | 0 | 2 ------- | -------- | -------- | -------- P(X) | 1/4 | 1/2 | 1/4

Step 5: Calculate the Mean of X

The mean (or expected value) of X, denoted by E(X), is calculated as: E(X) = Σ [x * P(x)] E(X) = (-2 * 1/4) + (0 * 1/2) + (2 * 1/4) E(X) = -2/4 + 0 + 2/4 E(X) = 0

Final Answer: The probability distribution is: X: -2, 0, 2 with probabilities 1/4, 1/2, 1/4 respectively. The mean is 0.

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