Answer
425.1k+ views
Hint:Probability of event to happen \[P\{ E\} \]
\[ = \dfrac{{Number\,of\,favourable\,outcome}}{{Total\,Number\,of\,outcomes}}\]
In this question we will be creating two different cases as the balls were drawn two times. In the first case the total number of balls in the bag were 6 out of which 3 are red and 3 are green. In the second case the number of balls in the bag are 6 as well out of which 3 are blue.
In the second case we don’t know anything about the colour of the other 3 balls those were drawn out from the bag.
Complete step-by- step solution:
Given the data according to the question and its probability cases
Case I:
There are 3 red and 3 green balls in a bag and a person draws 3 balls.
The possibilities are the following:
3 red, 0 green
2 red, 1 green
1 red, 2 green
0 red, 3 green
Case II:
As 3 blue balls have been added to the bag then the possibilities of the balls that are in the bag according to above cases are:
1) If 3 red, 0 green balls were drawn from the bag then there will be 0 red, 3 green and 3 blue balls in the bag.
2) 2 red, 1 green ball were drawn from the bag then there will be 1 red, 2 green and 3 blue balls in the bag.
3) 1 red, 2 green balls were drawn from the bag then there will be 2 red, 1 green and 3 blue balls in the bag.
4) 0 red, 3 green balls were drawn from the bag then there will be 3 red, 0 green and 3 blue balls in the bag.
We need to find the probability of three balls being different colours hence it rules out (a) and (d) cases.
Now, probability of withdrawing 2R, 1G is: \[ = \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
Probability of withdrawing 1R, 1G and 1B ball is: \[\dfrac{{{}^2{C_1} \times {}^1{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
Hence, the probability of both cases is:
\[ = \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}} \times \dfrac{{{}^2{C_1} \times {}^1{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
\[ = \dfrac{{\dfrac{{3!}}{{2!1!}} \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}} \times \dfrac{{\dfrac{{2!}}{{1!1!}} \times 1 \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}}\]
\[ = \dfrac{{\dfrac{{3 \times \not 2!}}{{\not 2!}} \times \dfrac{{3 \times \not 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{3! \times 3\not ! \times \not 2 \times 1}}}} \times \dfrac{{2! \times \dfrac{{3 \times 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{\not 3! \times 3! \times 2 \times 1}}}}\]
\[ = \dfrac{9}{{20}} \times \dfrac{6}{{20}} = \dfrac{{27}}{{200}}\,\,\,\,\,\,\,.......(1)\]
Now case II as the number of balls are same just colours are different, we have:
\[ = \dfrac{{{}^3{C_2} \times {}^3{C_1}}}{{{}^6{C_3}}} \times \dfrac{{{}^1{C_1} \times {}^2{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
\[ = \dfrac{{\dfrac{{3!}}{{2!1!}} \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}} \times \dfrac{{\dfrac{{2!}}{{1!1!}} \times 1 \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}}\]
\[ = \dfrac{{\dfrac{{3 \times \not 2!}}{{\not 2!}} \times \dfrac{{3 \times \not 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{3! \times 3\not ! \times \not 2 \times 1}}}} \times \dfrac{{2! \times \dfrac{{3 \times 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{\not 3! \times 3! \times 2 \times 1}}}}\]
\[ = \dfrac{9}{{20}} \times \dfrac{6}{{20}} = \dfrac{{27}}{{200}}\,\,\,\,\,\,\,.......(2)\]
\[ = \dfrac{{27}}{{200}}.....(2)\]
Probability of 3 later balls being all of different colours \[ = \dfrac{{27}}{{200}} + \dfrac{{27}}{{200}}\](from \[e{q^n}\] 1 & 2)
\[ = \dfrac{{2 \times 27}}{{200}}\]\[ = \dfrac{{27}}{{100}}\]
Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from \[I{I^{nd}}\].
In mathematics, a cube root of a number x is a number y such that \[\;{y^3}\; = \;x\]. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8, denoted \[\sqrt[3]{8}\], is 2, because \[{2^3}\; = \;8\]
\[ = \dfrac{{Number\,of\,favourable\,outcome}}{{Total\,Number\,of\,outcomes}}\]
In this question we will be creating two different cases as the balls were drawn two times. In the first case the total number of balls in the bag were 6 out of which 3 are red and 3 are green. In the second case the number of balls in the bag are 6 as well out of which 3 are blue.
In the second case we don’t know anything about the colour of the other 3 balls those were drawn out from the bag.
Complete step-by- step solution:
![seo images](https://www.vedantu.com/question-sets/6d534f87-e860-48df-b46e-d29d88bbd6e11953409089408122704.png)
Given the data according to the question and its probability cases
Case I:
There are 3 red and 3 green balls in a bag and a person draws 3 balls.
The possibilities are the following:
3 red, 0 green
2 red, 1 green
1 red, 2 green
0 red, 3 green
Case II:
As 3 blue balls have been added to the bag then the possibilities of the balls that are in the bag according to above cases are:
1) If 3 red, 0 green balls were drawn from the bag then there will be 0 red, 3 green and 3 blue balls in the bag.
2) 2 red, 1 green ball were drawn from the bag then there will be 1 red, 2 green and 3 blue balls in the bag.
3) 1 red, 2 green balls were drawn from the bag then there will be 2 red, 1 green and 3 blue balls in the bag.
4) 0 red, 3 green balls were drawn from the bag then there will be 3 red, 0 green and 3 blue balls in the bag.
We need to find the probability of three balls being different colours hence it rules out (a) and (d) cases.
Now, probability of withdrawing 2R, 1G is: \[ = \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
Probability of withdrawing 1R, 1G and 1B ball is: \[\dfrac{{{}^2{C_1} \times {}^1{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
Hence, the probability of both cases is:
\[ = \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}} \times \dfrac{{{}^2{C_1} \times {}^1{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
\[ = \dfrac{{\dfrac{{3!}}{{2!1!}} \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}} \times \dfrac{{\dfrac{{2!}}{{1!1!}} \times 1 \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}}\]
\[ = \dfrac{{\dfrac{{3 \times \not 2!}}{{\not 2!}} \times \dfrac{{3 \times \not 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{3! \times 3\not ! \times \not 2 \times 1}}}} \times \dfrac{{2! \times \dfrac{{3 \times 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{\not 3! \times 3! \times 2 \times 1}}}}\]
\[ = \dfrac{9}{{20}} \times \dfrac{6}{{20}} = \dfrac{{27}}{{200}}\,\,\,\,\,\,\,.......(1)\]
Now case II as the number of balls are same just colours are different, we have:
\[ = \dfrac{{{}^3{C_2} \times {}^3{C_1}}}{{{}^6{C_3}}} \times \dfrac{{{}^1{C_1} \times {}^2{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}\]
\[ = \dfrac{{\dfrac{{3!}}{{2!1!}} \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}} \times \dfrac{{\dfrac{{2!}}{{1!1!}} \times 1 \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}}\]
\[ = \dfrac{{\dfrac{{3 \times \not 2!}}{{\not 2!}} \times \dfrac{{3 \times \not 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{3! \times 3\not ! \times \not 2 \times 1}}}} \times \dfrac{{2! \times \dfrac{{3 \times 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{\not 3! \times 3! \times 2 \times 1}}}}\]
\[ = \dfrac{9}{{20}} \times \dfrac{6}{{20}} = \dfrac{{27}}{{200}}\,\,\,\,\,\,\,.......(2)\]
\[ = \dfrac{{27}}{{200}}.....(2)\]
Probability of 3 later balls being all of different colours \[ = \dfrac{{27}}{{200}} + \dfrac{{27}}{{200}}\](from \[e{q^n}\] 1 & 2)
\[ = \dfrac{{2 \times 27}}{{200}}\]\[ = \dfrac{{27}}{{100}}\]
Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from \[I{I^{nd}}\].
In mathematics, a cube root of a number x is a number y such that \[\;{y^3}\; = \;x\]. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8, denoted \[\sqrt[3]{8}\], is 2, because \[{2^3}\; = \;8\]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)