# 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.

Answer

Verified

368.4k+ views

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.

Last updated date: 29th Sep 2023

•

Total views: 368.4k

•

Views today: 9.68k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers