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Chapter 2: Problem 7

Suppose a coin having probability \(0.7\) of coming up heads is tossed three times. Let \(X\) denote the number of heads that appear in the three tosses. Determine the probability mass function of \(X\).

Short Answer

Expert verified
The probability mass function of the random variable X is given by: \[P(X = 0) = 0.027\] \[P(X = 1) = 0.189\] \[P(X = 2) = 0.441\] \[P(X = 3) = 0.343\]

Step by step solution

01

Determine possible values of X

X can take values from 0 to 3, representing the number of heads obtained in the three coin tosses: {0, 1, 2, 3}
02

Apply binomial probability formula for each value of X

Now we can apply the binomial probability formula with n = 3 and p = 0.7 for each value of X:
03

X = 0

\[P(X = 0) = \binom{3}{0} (0.7)^0 (0.3)^{3-0}\] \[P(X = 0) = 1 * 1 * (0.3)^3\] \[P(X = 0) = 0.027\]
04

X = 1

\[P(X = 1) = \binom{3}{1} (0.7)^1 (0.3)^{3-1}\] \[P(X = 1) = 3 * 0.7 * (0.3)^2\] \[P(X = 1) = 0.189\]
05

X = 2

\[P(X = 2) = \binom{3}{2} (0.7)^2 (0.3)^{3-2}\] \[P(X = 2) = 3 * 0.49 * 0.3\] \[P(X = 2) = 0.441\]
06

X = 3

\[P(X = 3) = \binom{3}{3} (0.7)^3 (0.3)^{3-3}\] \[P(X = 3) = 1 * (0.7)^3 * 1\] \[P(X = 3) = 0.343\]
07

Summarize the probability mass function

Now that we have the probabilities for each value of X, we can write the probability mass function using these values: \[P(X = 0) = 0.027\] \[P(X = 1) = 0.189\] \[P(X = 2) = 0.441\] \[P(X = 3) = 0.343\] So, with the provided probabilities, we have successfully determined the probability mass function of X.

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