1 of 8 0-day streak

Exact probability of repeated independent successes

Question 1 of 8

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A classification rule is applied independently to each record

The trials are independent. The probability of the target class on each record is \(\frac{5}{8}\), and the probability of another class is \(1-\frac{5}{8}\). The process is repeated for 7 records

Let \(X\) be the number of successes. What is the exact probability of 6 or more successes? Give an exact probability; an equivalent fraction or terminating decimal is acceptable

(a)1 mark

Which exact term set and calculation match \(\Pr(X\ge 6)\)?

Use the stated success counts, but do not count the possible positions of the successes separately

\left(\frac{5}{8}\right)^{6}\left(1-\frac{5}{8}\right)^{1} + \left(\frac{5}{8}\right)^{7}\left(1-\frac{5}{8}\right)^{0}

Add the terms for r = 6, 7

\binom{7}{6}\left(\frac{5}{8}\right)^{6}\left(1-\frac{5}{8}\right)^{1} + \binom{7}{7}\left(\frac{5}{8}\right)^{7}\left(1-\frac{5}{8}\right)^{0}

Subtract the event terms from 1: r = 6, 7

1 - (\binom{7}{6}\left(\frac{5}{8}\right)^{6}\left(1-\frac{5}{8}\right)^{1} + \binom{7}{7}\left(\frac{5}{8}\right)^{7}\left(1-\frac{5}{8}\right)^{0})

Use the stated success counts, with the same exponent on the success and failure factors

\binom{7}{6}\left(\frac{5}{8}\right)^{6}\left(1-\frac{5}{8}\right)^{6} + \binom{7}{7}\left(\frac{5}{8}\right)^{7}\left(1-\frac{5}{8}\right)^{7}

Use only the term for r = 6

\binom{7}{6}\left(\frac{5}{8}\right)^{6}\left(1-\frac{5}{8}\right)^{1}

(b)5 marks

For the representative term with \(r=6\), complete the exact formula data

n (number of trials) r (number of successes) C(n,r) success exponent failure exponent

Calculate \(\Pr(X\ge 6)\)

Exact probability

Format: exact terminating decimal