You are throwing darts at a dartboard. You have a [tex]\( \frac{1}{6} \)[/tex] chance of hitting the bull's-eye each time you throw. If you throw 3 times, what is the probability that you will hit the bull's-eye all 3 times?

A. [tex]\( \frac{5}{136} \)[/tex]

B. [tex]\( \frac{1}{218} \)[/tex]

C. [tex]\( \frac{3}{233} \)[/tex]

D. [tex]\( \frac{1}{100} \)[/tex]



Answer :

Let's solve the problem step-by-step.

First, we know that the probability of striking the bull's-eye in a single throw is [tex]\(\frac{1}{6}\)[/tex]. This means that for one throw, the chance of hitting the bull's-eye is [tex]\(\frac{1}{6}\)[/tex].

You want to find the probability of striking the bull's-eye three times in a row. When dealing with independent events (like multiple throws of a dart), the combined probability of all events occurring is the product of their individual probabilities.

Thus, the probability of striking the bull's-eye on the first throw, the second throw, and the third throw can be calculated by multiplying the probabilities for each throw together:

[tex]\[ \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) \][/tex]

This simplifies to:

[tex]\[ \left(\frac{1}{6}\right)^3 = \frac{1}{6^3} = \frac{1}{216} \][/tex]

Thus, the probability of striking the bull's-eye on all three throws is [tex]\(\frac{1}{216}\)[/tex].

Now let's verify if this matches one of the options provided:
A. [tex]\(\frac{5}{136}\)[/tex]
B. [tex]\(\frac{1}{218}\)[/tex]
C. [tex]\(\frac{3}{233}\)[/tex]
D. [tex]\(\frac{1}{100}\)[/tex]

The probability we calculated [tex]\(\frac{1}{216}\)[/tex] does not directly match any of the given options. So, we can confirm our detailed solution leads us to the correct answer that none of the options provided are entirely correct, strictly based on mathematical calculations. But if rounding or slight approximation needs to be considered, [tex]\(\frac{1}{218}\)[/tex] could be considered a very close evaluation.

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