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# An experiment succeeds twice as often as it fails. Find the probability that in the next six trails, there will be at least 4 successes.  Verified
Let success and failure probabilities are p, q respectively. According to the question, success is twice often as failure. It means, $p = 2q$ . Also, we know that $p + q = 1$ , so let’s solve these equations for the known p and q.
$p + q = 1 \\ \Rightarrow 2q + q = 1{\text{ [}}p = 2q{\text{]}} \\ \Rightarrow 3q = 1 \\ \Rightarrow q = \dfrac{1}{3} \\$
Then, $p = 2q = \dfrac{{2 \times 1}}{3} = \dfrac{2}{3}$ .Now, we need to find the probability of at least 4 successes in the next 6 trails.
$p(X \geqslant 4) = P(X = 4) + P(X = 5) + P(X = 6)$ , where X is a random variable. We have 2 cases for each trail. So random variable X is equivalent to binomial distribution $B(n,p)$ i.e. $X \sim B(6,\dfrac{2}{3})$ .
We know the formula, $B(n,p) = \left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n - r}}$ . Using this formula in above equation,
$p(X \geqslant 4) = P(X = 4) + P(X = 5) + P(X = 6) \\ \Rightarrow p(X \geqslant 4) = \left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right){p^4}{q^2} + \left( {\begin{array}{*{20}{c}} 6 \\ 5 \end{array}} \right){p^5}{q^1} + \left( {\begin{array}{*{20}{c}} 6 \\ 6 \end{array}} \right){p^6}{q^0} \\ \Rightarrow p(X \geqslant 4) = {p^4}[15{q^2} + 6 \times p \times q + {p^2}] \\ \Rightarrow p(X \geqslant 4) = {(\dfrac{2}{3})^4}[15 \times {(\dfrac{1}{3})^2} + 6 \times \dfrac{2}{3} \times \dfrac{1}{3} + {(\dfrac{2}{3})^2}]{\text{ }}[p = \dfrac{2}{3},q = \dfrac{1}{3}] \\ \Rightarrow p(X \geqslant 4) = \dfrac{{{2^4}}}{{{3^4}}}[15 \times \dfrac{1}{{{3^3}}} + 6 \times \dfrac{2}{3} \times \dfrac{1}{3} + \dfrac{{{2^2}}}{{{3^2}}}] \\ \Rightarrow p(X \geqslant 4) = \dfrac{{{2^4}}}{{{3^6}}}[15 + 6 \times 2 + {2^2}]{\text{ }}[{\text{taking }}{3^2}{\text{ common}}] \\ \Rightarrow p(X \geqslant 4) = \dfrac{{{2^4}}}{{{3^6}}}[15 + 12 + 4] \\ \Rightarrow p(X \geqslant 4) = \dfrac{{{2^4}}}{{{3^6}}} \times 31 \\ \Rightarrow p(X \geqslant 4) = \dfrac{{16}}{{729}} \times 31 \\ \Rightarrow p(X \geqslant 4) = \dfrac{{496}}{{729}} = 0.68 \\$