
On an average, rain falls on 12 days in every 30 days; find the probability that rain will fall on just 3 days of a given week.
Answer
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Hint: Use the binomial distribution formula\[q = 1 - p\], where \[q\]is the probability of failure, and \[p\]is the probability of success. The binomial distribution is the probability of success or failure outcome in an experiment or survey that is repeated multiple times.
Complete step-by-step answer:
In this question, we need to determine the probability with the condition that the rain will fall on just 3 days of a particular week for which we need to follow the formula of the Binomial theorem as \[P\left( r \right) = {}^n{C_r}{p^r}{q^{n - r}}\] where, ‘p’ is the probability of the success and ‘q’ is the probability of failure.
Average rainfalls on 12 days in every 30 days
Hence the probability that the rain falls on one day \[p = \dfrac{{12}}{{30}} = \dfrac{4}{{10}} = \dfrac{2}{5} - - - - (i)\]
Probability of failure of rainfall \[q = 1 - p = 1 - \dfrac{2}{5} = \dfrac{3}{5} - - - - (ii)\]
Since a week has 7 days, hence \[n = 7\]
The binomial theorem is given as \[P\left( r \right) = {}^n{C_r}{p^r}{q^{n - r}}\]
Since the probability of rainfall 3 days is required
Hence the required probability is \[P\left( {r \geqslant 3} \right) = 1 - P\left( {r \leqslant 2} \right)\]
This is equal to: \[P\left( {r \geqslant 3} \right) = 1 - [P\left( {r = 0} \right) + P\left( {r = 1} \right) + P\left( {r = 2} \right)]\]
Now using binomial theorem and from equation (i) and (ii), we can write
\[
P\left( {r \geqslant 3} \right) = 1 - [P\left( {r = 0} \right) + P\left( {r = 1} \right) + P\left( {r = 2} \right)] \\
= 1 - \left[ {{}^7{C_0}{{\left( {\dfrac{2}{3}} \right)}^7}{{\left( {\dfrac{1}{3}} \right)}^0} + {}^7{C_1}{{\left( {\dfrac{2}{3}} \right)}^6}{{\left( {\dfrac{1}{3}} \right)}^1} + {}^7{C_2}{{\left( {\dfrac{2}{3}} \right)}^5}{{\left( {\dfrac{1}{3}} \right)}^2}} \right] \\
= 1 - \left[ {\left( {1 \times \dfrac{{128}}{{2187}} \times 1} \right) + \left( {7 \times \dfrac{{64}}{{729}} \times \dfrac{1}{3}} \right) + \left( {21 \times \dfrac{{32}}{{243}} \times \dfrac{1}{9}} \right)} \right] \\
= 1 - \left( {\dfrac{{128}}{{2187}} + \dfrac{{448}}{{2187}} + \dfrac{{672}}{{2187}}} \right) \\
= 1 - \dfrac{{1248}}{{2187}} \\
= \dfrac{{939}}{{2187}} \\
= 0.43 \\
\]
Hence the probability that the rain will fall on just 3 days of a given week is 0.43.
Note: A binomial experiment is applicable when the experiment consists of ‘n’ identical trials; each trial results in two outcomes, either success or failure, the probability of success remains the same from trial to trial, and n trials are independent.
Complete step-by-step answer:
In this question, we need to determine the probability with the condition that the rain will fall on just 3 days of a particular week for which we need to follow the formula of the Binomial theorem as \[P\left( r \right) = {}^n{C_r}{p^r}{q^{n - r}}\] where, ‘p’ is the probability of the success and ‘q’ is the probability of failure.
Average rainfalls on 12 days in every 30 days
Hence the probability that the rain falls on one day \[p = \dfrac{{12}}{{30}} = \dfrac{4}{{10}} = \dfrac{2}{5} - - - - (i)\]
Probability of failure of rainfall \[q = 1 - p = 1 - \dfrac{2}{5} = \dfrac{3}{5} - - - - (ii)\]
Since a week has 7 days, hence \[n = 7\]
The binomial theorem is given as \[P\left( r \right) = {}^n{C_r}{p^r}{q^{n - r}}\]
Since the probability of rainfall 3 days is required
Hence the required probability is \[P\left( {r \geqslant 3} \right) = 1 - P\left( {r \leqslant 2} \right)\]
This is equal to: \[P\left( {r \geqslant 3} \right) = 1 - [P\left( {r = 0} \right) + P\left( {r = 1} \right) + P\left( {r = 2} \right)]\]
Now using binomial theorem and from equation (i) and (ii), we can write
\[
P\left( {r \geqslant 3} \right) = 1 - [P\left( {r = 0} \right) + P\left( {r = 1} \right) + P\left( {r = 2} \right)] \\
= 1 - \left[ {{}^7{C_0}{{\left( {\dfrac{2}{3}} \right)}^7}{{\left( {\dfrac{1}{3}} \right)}^0} + {}^7{C_1}{{\left( {\dfrac{2}{3}} \right)}^6}{{\left( {\dfrac{1}{3}} \right)}^1} + {}^7{C_2}{{\left( {\dfrac{2}{3}} \right)}^5}{{\left( {\dfrac{1}{3}} \right)}^2}} \right] \\
= 1 - \left[ {\left( {1 \times \dfrac{{128}}{{2187}} \times 1} \right) + \left( {7 \times \dfrac{{64}}{{729}} \times \dfrac{1}{3}} \right) + \left( {21 \times \dfrac{{32}}{{243}} \times \dfrac{1}{9}} \right)} \right] \\
= 1 - \left( {\dfrac{{128}}{{2187}} + \dfrac{{448}}{{2187}} + \dfrac{{672}}{{2187}}} \right) \\
= 1 - \dfrac{{1248}}{{2187}} \\
= \dfrac{{939}}{{2187}} \\
= 0.43 \\
\]
Hence the probability that the rain will fall on just 3 days of a given week is 0.43.
Note: A binomial experiment is applicable when the experiment consists of ‘n’ identical trials; each trial results in two outcomes, either success or failure, the probability of success remains the same from trial to trial, and n trials are independent.
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