# A bag contains 7 tickets marked with the numbers 0,1, 2….6 respectively. A ticket is drawn and replaced; find the chance that after 4 drawings the sum of the numbers drawn is 8.

Last updated date: 19th Mar 2023

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Answer

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Hint: Use permutations to find the number of favourable cases.

Given the tickets are marked with numbers 0 to 6.

We are allowed to draw 4 times.

First, we have to find the number of ways to draw 4 tickets such that the sum of the numbers on the ticket is equal to 8.

We can choose the 4 numbers from 0 to 6 having sum equal to 8 with repetition and then use permutations to find the number of favourable ways in each case.

It is to note that the numbers can be repeated as a ticket is replaced as soon as it is drawn from the bag.

So, the required combinations and the corresponding ways are:

$

6{\text{ 2 0 0 ; }}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

6{\text{ 1 0 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{5 3 0 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{5 2 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{5 1 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{3!}} = 4 \\

{\text{4 4 0 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!2!}} = 6 \\

{\text{4 3 1 0 ;}}no.{\text{ }}of{\text{ }}ways = 4! = 24 \\

{\text{4 2 2 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{4 2 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{3 3 2 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{3 3 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!2!}} = 6 \\

{\text{3 2 2 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{2 2 2 2 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{4!}} = 1 \\

$

Total number of favourable cases $n\left( E \right) = 149$

Since each draw can have 7 possible numbers on it.

Therefore, total number of ways in which 4 tickets could be drawn from the bag with replacement,$n\left( S \right) = {7^4}$

Probability or chance that after 4 drawings the sum of the numbers drawn is 8,$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{{149}}{{{7^4}}} = \dfrac{{149}}{{2401}} \cong 0.062$

Hence the chance of getting sum 8 after 4 drawings is approximately 0.062.

Note: Try to solve the probability problems with a large sample space using permutation and combination. While counting the favourable cases in the above problem, the numbers could be chosen such that the number on the left is always greater than or equal to the number on its right, not less. This approach could help in saving time and rectifying the redundant cases.

Given the tickets are marked with numbers 0 to 6.

We are allowed to draw 4 times.

First, we have to find the number of ways to draw 4 tickets such that the sum of the numbers on the ticket is equal to 8.

We can choose the 4 numbers from 0 to 6 having sum equal to 8 with repetition and then use permutations to find the number of favourable ways in each case.

It is to note that the numbers can be repeated as a ticket is replaced as soon as it is drawn from the bag.

So, the required combinations and the corresponding ways are:

$

6{\text{ 2 0 0 ; }}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

6{\text{ 1 0 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{5 3 0 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{5 2 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{5 1 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{3!}} = 4 \\

{\text{4 4 0 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!2!}} = 6 \\

{\text{4 3 1 0 ;}}no.{\text{ }}of{\text{ }}ways = 4! = 24 \\

{\text{4 2 2 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{4 2 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{3 3 2 0 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{3 3 1 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!2!}} = 6 \\

{\text{3 2 2 1 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{2!}} = 12 \\

{\text{2 2 2 2 ;}}no.{\text{ }}of{\text{ }}ways = \dfrac{{4!}}{{4!}} = 1 \\

$

Total number of favourable cases $n\left( E \right) = 149$

Since each draw can have 7 possible numbers on it.

Therefore, total number of ways in which 4 tickets could be drawn from the bag with replacement,$n\left( S \right) = {7^4}$

Probability or chance that after 4 drawings the sum of the numbers drawn is 8,$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{{149}}{{{7^4}}} = \dfrac{{149}}{{2401}} \cong 0.062$

Hence the chance of getting sum 8 after 4 drawings is approximately 0.062.

Note: Try to solve the probability problems with a large sample space using permutation and combination. While counting the favourable cases in the above problem, the numbers could be chosen such that the number on the left is always greater than or equal to the number on its right, not less. This approach could help in saving time and rectifying the redundant cases.

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