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$\left( i \right)$ All are blue?

$\left( {ii} \right)$ At least one will be green?

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Hint: - The question can be solved easily by the use of combination theory of selection of items and then using basic definition of probability.

$\left( i \right)$ All will be blue.

As we know that if we have to select $n$ items from $m$items then according to combination theory it can be done in ${}^m{C_n}$ different ways.

Given that 5 marbles are to be chosen from 60 marbles (10 red, 20 blue and 30 green).

Hence, no of ways of selection ${\text{n}}\left( S \right) = {}^{60}{C_5}$

Let $A$ be the event that all are blue.

Out of 20 blue marbles, 5 will be drawn. Hence, No of ways of such selection is:

$n\left( A \right) = {}^{20}{C_5}$

Since by the basic definition of probability we know that

P(A)=favorable cases/total cases

$

\therefore P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{{{}^{20}{C_5}}}{{{}^{60}{C_5}}} \\

= \dfrac{{\dfrac{{20!}}{{5! \times 15!}}}}{{\dfrac{{60!}}{{5! \times 55!}}}}\left[ {\because {}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}} \right] \\

= \dfrac{{\dfrac{{20!}}{{15!}}}}{{\dfrac{{60!}}{{55!}}}} = \dfrac{{20! \times 55!}}{{15! \times 60!}} \\

$

$\left( {ii} \right)$ At least one will be green.

$P\left( {{\text{atleast one will be green}}} \right) = 1 - P\left( {{\text{none of the marbles are green}}} \right)$

Let’s first calculate the probability that none of the drawn marbles are green.

Let B be the event that none of the marbles are green which makes the total available marbles = 60-30 = 30

Now, out of these 30 marbles, 5 are to be drawn.

Hence, No of ways of such selection is:

Hence, $n\left( B \right) = {}^{30}{C_5}$

By the basic definition of probability we know that:

\[

P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{{{}^{30}{C_5}}}{{{}^{60}{C_5}}} \\

\therefore P\left( {{\text{at least one will be green}}} \right) = 1 - P\left( {{\text{none of the marbles are green}}} \right) \\

{\text{ }} = 1 - \dfrac{{{}^{30}{C_5}}}{{{}^{60}{C_5}}} \\

\Rightarrow P\left( {{\text{at least one will be green}}} \right) = \dfrac{{{}^{60}{C_5} - {}^{30}{C_5}}}{{{}^{60}{C_5}}} \\

= \dfrac{{\dfrac{{60!}}{{5! \times 55!}} - \dfrac{{30!}}{{5! \times 25!}}}}{{\dfrac{{60!}}{{5! \times 55!}}}}\left[ {\because {}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}} \right] \\

= \dfrac{{\dfrac{{60!}}{{55!}} - \dfrac{{30!}}{{25!}}}}{{\dfrac{{60!}}{{55!}}}} \\

\]

Note: Using the method of combination is one of the best and shortest ways to find the probability but we need to take extra caution on the events that are mutually exclusive. Also, in some cases where we are unable to find the probability of happening of an event directly, we rather calculate the probability of not happening of that event and further subtract it by one.

$\left( i \right)$ All will be blue.

As we know that if we have to select $n$ items from $m$items then according to combination theory it can be done in ${}^m{C_n}$ different ways.

Given that 5 marbles are to be chosen from 60 marbles (10 red, 20 blue and 30 green).

Hence, no of ways of selection ${\text{n}}\left( S \right) = {}^{60}{C_5}$

Let $A$ be the event that all are blue.

Out of 20 blue marbles, 5 will be drawn. Hence, No of ways of such selection is:

$n\left( A \right) = {}^{20}{C_5}$

Since by the basic definition of probability we know that

P(A)=favorable cases/total cases

$

\therefore P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{{{}^{20}{C_5}}}{{{}^{60}{C_5}}} \\

= \dfrac{{\dfrac{{20!}}{{5! \times 15!}}}}{{\dfrac{{60!}}{{5! \times 55!}}}}\left[ {\because {}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}} \right] \\

= \dfrac{{\dfrac{{20!}}{{15!}}}}{{\dfrac{{60!}}{{55!}}}} = \dfrac{{20! \times 55!}}{{15! \times 60!}} \\

$

$\left( {ii} \right)$ At least one will be green.

$P\left( {{\text{atleast one will be green}}} \right) = 1 - P\left( {{\text{none of the marbles are green}}} \right)$

Let’s first calculate the probability that none of the drawn marbles are green.

Let B be the event that none of the marbles are green which makes the total available marbles = 60-30 = 30

Now, out of these 30 marbles, 5 are to be drawn.

Hence, No of ways of such selection is:

Hence, $n\left( B \right) = {}^{30}{C_5}$

By the basic definition of probability we know that:

\[

P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{{{}^{30}{C_5}}}{{{}^{60}{C_5}}} \\

\therefore P\left( {{\text{at least one will be green}}} \right) = 1 - P\left( {{\text{none of the marbles are green}}} \right) \\

{\text{ }} = 1 - \dfrac{{{}^{30}{C_5}}}{{{}^{60}{C_5}}} \\

\Rightarrow P\left( {{\text{at least one will be green}}} \right) = \dfrac{{{}^{60}{C_5} - {}^{30}{C_5}}}{{{}^{60}{C_5}}} \\

= \dfrac{{\dfrac{{60!}}{{5! \times 55!}} - \dfrac{{30!}}{{5! \times 25!}}}}{{\dfrac{{60!}}{{5! \times 55!}}}}\left[ {\because {}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}} \right] \\

= \dfrac{{\dfrac{{60!}}{{55!}} - \dfrac{{30!}}{{25!}}}}{{\dfrac{{60!}}{{55!}}}} \\

\]

Note: Using the method of combination is one of the best and shortest ways to find the probability but we need to take extra caution on the events that are mutually exclusive. Also, in some cases where we are unable to find the probability of happening of an event directly, we rather calculate the probability of not happening of that event and further subtract it by one.