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Question

Answers

a) \[1-\left\{ \dfrac{20+10N}{{}^{N+7}{{C}_{5}}} \right\}\]

b) \[\left\{ \dfrac{20+10N}{{}^{N+7}{{C}_{5}}} \right\}\]

c) \[\left\{ \dfrac{20+10N}{{}^{N+5}{{C}_{7}}} \right\}\]

d) \[1-\left\{ \dfrac{20+10N}{{}^{N+5}{{C}_{7}}} \right\}\]

Answer
Verified

Hint: Find the total number of coins present in the box. Select 5 coins from the total number of coins. Find the favorable conditions of how 5 coins can be selected. Then find P (E).

Complete step-by-step answer:

From the question, a box contains coins. It is mentioned that these are 4 types of coins, fifty paise, twenty paise, ten paise and five paise.

There are 2 fifty paise coins, 5 twenty paise coins and the rest are ten and five paise coins, where the number can be taken as N, where N is a natural number.

Therefore the total number of coins = N + 7.

2 fifty paise + 5 twenty paise + N = N + 7.

We have to take 5 coins from the box in random i.e. we have to take 5 coins from the (N + 7) coins present in the box.

\[\therefore \]5 coins can be chosen from (N + 7) coins in \[{}^{N+7}{{C}_{5}}\]ways, which is of the form \[{}^{n}{{C}_{r}}\]of combination, where no order is needed for choosing the 5 coins.

Let E denote the event that the sum of the values of the coins is less than 1 rupee and fifty paise.

Let Eâ€™ denotes the event that the total value of five coins is equal to or more than 1 rupee and 50 paise.

There are a number of favorable cases, or which we have to choose 5 coins.

1 coin from 50 paise and 4 coins from 20 paise\[\Rightarrow {}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}\]

2 coins from 50 paise and 3 coins from 20 paise \[\Rightarrow {}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}\]

2 coins from 50 paise, 2 coins from 20 paise and 1 from N\[\Rightarrow {}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}\]

There 3 are the favorable cases, let us add them together.

\[{}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}+{}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}+{}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}\]

We know, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]

\[\left[ \begin{align}

& \because {}^{N}{{C}_{0}}=1 \\

& \because {}^{2}{{C}_{2}}=1 \\

\end{align} \right]\]

\[\begin{align}

& \therefore \dfrac{2!}{1!1!}\times \dfrac{5!}{1!\times 4!}\times 1+\dfrac{2!}{0!1!}\times \dfrac{5!}{2!3!}\times 1+\dfrac{2!}{0!1!}\times \dfrac{5!}{2!3!}\times N \\

& =2\times 5\times 1+1\times \dfrac{5\times 4\times 3!}{2!3!}\times 1+1\times \dfrac{5\times 4\times 3!}{2!31}\times N \\

& =10+10\times 1+10\times N \\

& =10+10+10N \\

& =20+10N \\

& =10\left( 2+N \right) \\

\end{align}\]

\[P\left( E \right)=10\left( N+2 \right)/{}^{N+7}{{C}_{5}}\]; Probability of sum of value of coin less than Rs. 1 and 50 paise.

We know, P (E) + P (not E) =1

\[\begin{align}

& P\left( E \right)+P\left( E' \right)=1 \\

& \therefore P\left( E' \right)=1-P\left( E \right)=1-\dfrac{10\left( N+2 \right)}{{}^{N+7}{{C}_{5}}} \\

\end{align}\]

\[\therefore \]Required probability \[=\dfrac{{}^{n+7}{{C}_{5}}-10\left( N+2 \right)}{{}^{n+7}{{C}_{5}}}\]

\[=1-\dfrac{10\left( N+2 \right)}{{}^{n+7}{{C}_{5}}}=1-\left\{ \dfrac{20+10N}{{}^{n+7}{{C}_{5}}} \right\}\]

Hence option (a) is the correct answer.

Note: In questions like their basic understanding of the problem is required. The total number of coins formed is \[\left( N+7 \right)\]. But we will find it difficult to understand how many coins were there in the box. You have to pick 5 coins randomly from a box consisting of 4 different sets of paisa. Find the favorable conditions for the same.

Complete step-by-step answer:

From the question, a box contains coins. It is mentioned that these are 4 types of coins, fifty paise, twenty paise, ten paise and five paise.

There are 2 fifty paise coins, 5 twenty paise coins and the rest are ten and five paise coins, where the number can be taken as N, where N is a natural number.

Therefore the total number of coins = N + 7.

2 fifty paise + 5 twenty paise + N = N + 7.

We have to take 5 coins from the box in random i.e. we have to take 5 coins from the (N + 7) coins present in the box.

\[\therefore \]5 coins can be chosen from (N + 7) coins in \[{}^{N+7}{{C}_{5}}\]ways, which is of the form \[{}^{n}{{C}_{r}}\]of combination, where no order is needed for choosing the 5 coins.

Let E denote the event that the sum of the values of the coins is less than 1 rupee and fifty paise.

Let Eâ€™ denotes the event that the total value of five coins is equal to or more than 1 rupee and 50 paise.

There are a number of favorable cases, or which we have to choose 5 coins.

1 coin from 50 paise and 4 coins from 20 paise\[\Rightarrow {}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}\]

2 coins from 50 paise and 3 coins from 20 paise \[\Rightarrow {}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}\]

2 coins from 50 paise, 2 coins from 20 paise and 1 from N\[\Rightarrow {}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}\]

There 3 are the favorable cases, let us add them together.

\[{}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}+{}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}+{}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}\]

We know, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]

\[\left[ \begin{align}

& \because {}^{N}{{C}_{0}}=1 \\

& \because {}^{2}{{C}_{2}}=1 \\

\end{align} \right]\]

\[\begin{align}

& \therefore \dfrac{2!}{1!1!}\times \dfrac{5!}{1!\times 4!}\times 1+\dfrac{2!}{0!1!}\times \dfrac{5!}{2!3!}\times 1+\dfrac{2!}{0!1!}\times \dfrac{5!}{2!3!}\times N \\

& =2\times 5\times 1+1\times \dfrac{5\times 4\times 3!}{2!3!}\times 1+1\times \dfrac{5\times 4\times 3!}{2!31}\times N \\

& =10+10\times 1+10\times N \\

& =10+10+10N \\

& =20+10N \\

& =10\left( 2+N \right) \\

\end{align}\]

\[P\left( E \right)=10\left( N+2 \right)/{}^{N+7}{{C}_{5}}\]; Probability of sum of value of coin less than Rs. 1 and 50 paise.

We know, P (E) + P (not E) =1

\[\begin{align}

& P\left( E \right)+P\left( E' \right)=1 \\

& \therefore P\left( E' \right)=1-P\left( E \right)=1-\dfrac{10\left( N+2 \right)}{{}^{N+7}{{C}_{5}}} \\

\end{align}\]

\[\therefore \]Required probability \[=\dfrac{{}^{n+7}{{C}_{5}}-10\left( N+2 \right)}{{}^{n+7}{{C}_{5}}}\]

\[=1-\dfrac{10\left( N+2 \right)}{{}^{n+7}{{C}_{5}}}=1-\left\{ \dfrac{20+10N}{{}^{n+7}{{C}_{5}}} \right\}\]

Hence option (a) is the correct answer.

Note: In questions like their basic understanding of the problem is required. The total number of coins formed is \[\left( N+7 \right)\]. But we will find it difficult to understand how many coins were there in the box. You have to pick 5 coins randomly from a box consisting of 4 different sets of paisa. Find the favorable conditions for the same.

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