1 of 8 0-day streak

Exact probability of repeated independent successes

Question 1 of 8

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A laboratory procedure is repeated independently for each sample

The trials are independent. The probability of the specified result on each run is \(\frac{7}{10}\), and the probability of another result is \(1-\frac{7}{10}\). The process is repeated for 7 runs

Let \(X\) be the number of successes. Calculate the probability that the number of successes is at least 6. Give an exact probability; an equivalent fraction or terminating decimal is acceptable

(a)1 mark

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

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

1 - (\binom{7}{6}\left(\frac{7}{10}\right)^{6}\left(1-\frac{7}{10}\right)^{1} + \binom{7}{7}\left(\frac{7}{10}\right)^{7}\left(1-\frac{7}{10}\right)^{0})

Add the terms for r = 6, 7

\binom{7}{6}\left(\frac{7}{10}\right)^{6}\left(1-\frac{7}{10}\right)^{1} + \binom{7}{7}\left(\frac{7}{10}\right)^{7}\left(1-\frac{7}{10}\right)^{0}

Use only the term for r = 6

\binom{7}{6}\left(\frac{7}{10}\right)^{6}\left(1-\frac{7}{10}\right)^{1}

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

\binom{7}{6}\left(\frac{7}{10}\right)^{6}\left(1-\frac{7}{10}\right)^{6} + \binom{7}{7}\left(\frac{7}{10}\right)^{7}\left(1-\frac{7}{10}\right)^{7}

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

\left(\frac{7}{10}\right)^{6}\left(1-\frac{7}{10}\right)^{1} + \left(\frac{7}{10}\right)^{7}\left(1-\frac{7}{10}\right)^{0}

(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