Answer
Verified
480.9k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE