Math Problem Statement

An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc.

For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hth, then N (hth) = 1. Suppose that the random variable X is defined in terms of N as follows: X = N2-3N-3. The values of X are given in the table below.

Outcome

tth

thh

hht

tht hhh

itt

het hth

Value of

X-5-5-5-5-3-3-5-5

Calculate the probabilities P(X= x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.

Value x of X

P(X=x)

X

5

Solution

To solve this problem, let's break it down step by step.

Step 1: Understand the Random Variable XX

The random variable XX is defined in terms of NN, which is the number of tails in a particular outcome. The function defining XX is given by: X=N23N3X = N^2 - 3N - 3

Step 2: Determine NN for Each Outcome

List the possible outcomes when a fair coin is tossed 3 times. There are 23=82^3 = 8 possible outcomes:

  1. hhhhhh (0 tails)
  2. hhthht (1 tail)
  3. hthhth (1 tail)
  4. thhthh (1 tail)
  5. htthtt (2 tails)
  6. thttht (2 tails)
  7. tthtth (2 tails)
  8. tttttt (3 tails)

For each outcome, count the number of tails NN and calculate XX using the formula X=N23N3X = N^2 - 3N - 3.

Step 3: Calculate the Values of XX

Now, calculate XX for each possible value of NN:

  • N=0N = 0: X=023(0)3=3X = 0^2 - 3(0) - 3 = -3
  • N=1N = 1: X=123(1)3=133=5X = 1^2 - 3(1) - 3 = 1 - 3 - 3 = -5
  • N=2N = 2: X=223(2)3=463=5X = 2^2 - 3(2) - 3 = 4 - 6 - 3 = -5
  • N=3N = 3: X=323(3)3=993=3X = 3^2 - 3(3) - 3 = 9 - 9 - 3 = -3

Step 4: Assign XX Values to Each Outcome

Using the outcomes listed above, assign the calculated XX values:

  • hhhhhh (0 tails): X=3X = -3
  • hhthht (1 tail): X=5X = -5
  • hthhth (1 tail): X=5X = -5
  • thhthh (1 tail): X=5X = -5
  • htthtt (2 tails): X=5X = -5
  • thttht (2 tails): X=5X = -5
  • tthtth (2 tails): X=5X = -5
  • tttttt (3 tails): X=3X = -3

Step 5: Calculate the Probability Distribution P(X=x)P(X = x)

Now, determine the probability of each value of XX occurring.

  • X=3X = -3 occurs in 2 outcomes: hhhhhh and tttttt.
  • X=5X = -5 occurs in 6 outcomes: hhthht, hthhth, thhthh, htthtt, thttht, and tthtth.

Since each outcome is equally likely with probability 18\frac{1}{8}:

P(X=3)=28=14P(X = -3) = \frac{2}{8} = \frac{1}{4} P(X=5)=68=34P(X = -5) = \frac{6}{8} = \frac{3}{4}

Final Answer

The probability distribution of XX is as follows:

X & -5 & -3 \\ \hline P(X = x) & \frac{3}{4} & \frac{1}{4} \\ \end{array}$$ Would you like more detailed steps or explanations about this problem? Here are some related questions: 1. How do you compute the expected value of $$X$$ based on this probability distribution? 2. What is the variance of $$X$$ in this scenario? 3. How do you interpret the random variable $$X$$ in this context? 4. How does the binomial distribution relate to this problem? 5. How would the distribution change if the coin was biased? 6. Can you explain the concept of conditional probability using this example? 7. What is the significance of the function $$X = N^2 - 3N - 3$$? 8. How do these results change if the number of coin tosses is increased? **Tip**: When working with probability distributions, always ensure that the sum of the probabilities equals 1 as a check for your calculations.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Random Variables
Expected Value

Formulas

X = N^2 - 3N - 3

Theorems

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Suitable Grade Level

College Level