Question

An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.

 

Answer 1

Let X denote the number of successes in 6 trials.

\[\text{ It is given that successes are twice the failures }. \]
\[ \Rightarrow p = 2q\]
\[ p + q = 1\]
\[ \Rightarrow 3q = 1\]
\[ \Rightarrow q = \frac{1}{3}\]
\[ \therefore p = 1 - \frac{1}{3} = \frac{2}{3}\]
\[ n = 6\]
\[\text{ Hence, the distribution is given by } \]

\[P(X = r) =^{6}{}{C}_r \left( \frac{2}{3} \right)^r \left( \frac{1}{3} \right)^{6 - r} , r = 0, 1, 2 . . . . . 6\]
\[P(\text{ atleast 4 successes} ) = P(X \geq 4) \]
\[ = P(X = 4) + P(X = 5) + P(X = 6)\]
\[^{6}{}{C}_4 \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^{6 - 4} + ^{6}{}{C}_5 \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^{6 - 6} \]
\[ = \frac{15( 2^4 ) + 6(32) + 64}{3^6}\]
\[ = \frac{240 + 192 + 64}{729}\]
\[ = \frac{496}{729}\]