Answer
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Hint: In this question, we need to determine the probability for at least four swimmers out of five and at most three swimmers out of five such that the probability that a student cannot swim is 0.20. For this, we will use the concept of dependent events for the given conditions in the question. By applying the formula of binomial distribution we arrive at the required result.
Complete step-by-step answer:
The probability that the student is not a swimmer is $ \dfrac{1}{5} $ or $ p = \dfrac{1}{5} $
So, the probability that a student will be a swimmer is given\[\left( {1 - 0.2} \right) = 0.8\] or, $ q = \dfrac{4}{5} $
(i) The probability of having four swimmers out of five students is given as:
$ P(4) = {}^5{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^1} $
Again, the probability of having five swimmers out of five students is given as:
$ P(5) = {}^5{C_5}{\left( {\dfrac{4}{5}} \right)^5}{\left( {\dfrac{1}{5}} \right)^0} $
Hence, the probability of having at least four swimmers is the summation of the probability of having four swimmers and five swimmers out of five swimmers.
$
\Rightarrow P(4,5) = P(4) + P(5) \\
= {}^5{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^1} + {}^5{C_5}{\left( {\dfrac{4}{5}} \right)^5}{\left( {\dfrac{1}{5}} \right)^0} \\
= 5{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^1} + {\left( {\dfrac{4}{5}} \right)^5} \\
= {\left( {\dfrac{4}{5}} \right)^4}\left( {1 + \dfrac{4}{5}} \right) \\
= \dfrac{{256}}{{625}} \times \dfrac{9}{5} \\
= 0.73728 \\
$
Hence, the probability of having at least four swimmers out of five students is 0.73728.
(ii)The probability of having at most three swimmers out of five swimmers is the summation of the probability of having no swimmer, one swimmer, two swimmers and three swimmers. Mathematically,
$ P(0,1,2,3) = P(0) + P(1) + P(2) + P(3) $
Or, this can also be written as:
$ \Rightarrow P(0,1,2,3) = 1 - P(4,5) $
So, substituting the value of P(4,5) in the above equation, we get
$
P(0,1,2,3) = 1 - 0.73728 \\
= 0.26272 \\
$
Hence, the probability of having at most three swimmers out of five students is 0.26272.
Note: It is worth noting down here that the term at least means “a minimum count of” whereas the term at most implies “a maximum count of''. Here, in the question at least four swimmers means that four and five swimmers out of a total five students whereas at most three implies no swimmer, one swimmer, two swimmers and three swimmers.
Complete step-by-step answer:
The probability that the student is not a swimmer is $ \dfrac{1}{5} $ or $ p = \dfrac{1}{5} $
So, the probability that a student will be a swimmer is given\[\left( {1 - 0.2} \right) = 0.8\] or, $ q = \dfrac{4}{5} $
(i) The probability of having four swimmers out of five students is given as:
$ P(4) = {}^5{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^1} $
Again, the probability of having five swimmers out of five students is given as:
$ P(5) = {}^5{C_5}{\left( {\dfrac{4}{5}} \right)^5}{\left( {\dfrac{1}{5}} \right)^0} $
Hence, the probability of having at least four swimmers is the summation of the probability of having four swimmers and five swimmers out of five swimmers.
$
\Rightarrow P(4,5) = P(4) + P(5) \\
= {}^5{C_4}{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^1} + {}^5{C_5}{\left( {\dfrac{4}{5}} \right)^5}{\left( {\dfrac{1}{5}} \right)^0} \\
= 5{\left( {\dfrac{4}{5}} \right)^4}{\left( {\dfrac{1}{5}} \right)^1} + {\left( {\dfrac{4}{5}} \right)^5} \\
= {\left( {\dfrac{4}{5}} \right)^4}\left( {1 + \dfrac{4}{5}} \right) \\
= \dfrac{{256}}{{625}} \times \dfrac{9}{5} \\
= 0.73728 \\
$
Hence, the probability of having at least four swimmers out of five students is 0.73728.
(ii)The probability of having at most three swimmers out of five swimmers is the summation of the probability of having no swimmer, one swimmer, two swimmers and three swimmers. Mathematically,
$ P(0,1,2,3) = P(0) + P(1) + P(2) + P(3) $
Or, this can also be written as:
$ \Rightarrow P(0,1,2,3) = 1 - P(4,5) $
So, substituting the value of P(4,5) in the above equation, we get
$
P(0,1,2,3) = 1 - 0.73728 \\
= 0.26272 \\
$
Hence, the probability of having at most three swimmers out of five students is 0.26272.
Note: It is worth noting down here that the term at least means “a minimum count of” whereas the term at most implies “a maximum count of''. Here, in the question at least four swimmers means that four and five swimmers out of a total five students whereas at most three implies no swimmer, one swimmer, two swimmers and three swimmers.
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