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consider an experiment in which 13 identical small boxes are placed sid…

Question

consider an experiment in which 13 identical small boxes are placed side by side on a table. a crystal is placed at random inside one of the boxes. a self - professed \psychic\ is asked to pick the box that contains the crystal. complete parts a through c. a. if the \psychic\ simply guesses, what is the probability that she picks the box with the crystal? the probability is $\frac{1}{13}$. (type an exact answer in simplified form.) b. if the experiment is repeated six times, what is the probability that the \psychic\ guesses correctly at least once? the probability is . (round to the nearest thousandth as needed.)

Explanation:

Step1: Calculate probability of single - guess

There are 13 boxes and 1 has the crystal. The probability of guessing correctly in a single try, $p=\frac{1}{13}$, and the probability of guessing incorrectly, $q = 1 - p=1-\frac{1}{13}=\frac{12}{13}$.

Step2: Calculate probability of never guessing correctly in 6 tries

The probability of never guessing correctly in $n = 6$ independent tries is $q^n$. Using the formula for independent events, we have $(\frac{12}{13})^6=\frac{12^6}{13^6}=\frac{2985984}{4826809}$.

Step3: Calculate probability of guessing correctly at least once in 6 tries

The probability of guessing correctly at least once in 6 tries is the complement of never guessing correctly. Let $P(X\geq1)$ be the probability of guessing correctly at least once. Then $P(X\geq1)=1 - P(X = 0)$. So $P(X\geq1)=1-(\frac{12}{13})^6=1-\frac{2985984}{4826809}=\frac{4826809 - 2985984}{4826809}=\frac{1840825}{4826809}\approx0.381$.

Answer:

a. $\frac{1}{13}$
b. $0.381$