Answer

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Hint: Use binomial distribution.

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.

Since it's at least so it can be more than that also. Mathematically,

$p(X \geqslant 4) = P(X = 4) + P(X = 5) + P(X = 6)$ , 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 \\

\]

Hence, the required probability is 0.68.

Note: These kinds of problems are core in probability. First with success and failure conditions we got the probabilities. Then, the hack is understanding the binomial distribution of random variables. As the word says, bi-nomial. whenever we have two cases like win or lose, then immediately binomial distribution should come in our mind. After getting it, a simple distribution formula will lead us to the answer.

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.

Since it's at least so it can be more than that also. Mathematically,

$p(X \geqslant 4) = P(X = 4) + P(X = 5) + P(X = 6)$ , 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 \\

\]

Hence, the required probability is 0.68.

Note: These kinds of problems are core in probability. First with success and failure conditions we got the probabilities. Then, the hack is understanding the binomial distribution of random variables. As the word says, bi-nomial. whenever we have two cases like win or lose, then immediately binomial distribution should come in our mind. After getting it, a simple distribution formula will lead us to the answer.

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