
An urn contains 5 red , 4 black and 3 white marbles. Then the number of ways in which marbles can be drawn from it so that at most 3 of them are red, is :
E.455
F.460
G.490
H.495
Answer
485.7k+ views
Hint:
In this question given that urn has 5 Red, 4 black and 3 White marbles. And we draw 3 balls in a single draw so that at most 3 are red. At most 3 red balls either contain zero red ball, one red ball, 2 red ball or 3 red ball. We can either calculate this by calculating a single case every time. Or we can proceed it by other ways like calculation total selected – no of ways to select 4 balls.
Complete step by step solution:
A urn contains 5 red , 4 black and 3 white balls . And in a single draw we get at most 3 red balls.
We have to find the no of ways to select at most 3 red ball
= ( 0 Red ball + other 3 balls ) + ( 1 Red ball + other 3 balls) + ( 2 Red ball + other 2 balls )
+ ( 3 Red ball + 1 other )
= no of ways to select 4 balls – no of select 4 red balls
\[ = {}^{12}{C_4} - {}^5{C_4}\]
\[\begin{array}{l}
= \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}} - 5\\
= 495 - 5\\
= 490
\end{array}\]
Hence the required number of ways is 490.
Note:
In this type of question students generally confuse at most and at least , more than and less than so take care of it during attempting these types of questions . Almost means max you get 3 red ball in a draw or minimum you can get zero
In this question given that urn has 5 Red, 4 black and 3 White marbles. And we draw 3 balls in a single draw so that at most 3 are red. At most 3 red balls either contain zero red ball, one red ball, 2 red ball or 3 red ball. We can either calculate this by calculating a single case every time. Or we can proceed it by other ways like calculation total selected – no of ways to select 4 balls.
Complete step by step solution:
A urn contains 5 red , 4 black and 3 white balls . And in a single draw we get at most 3 red balls.
We have to find the no of ways to select at most 3 red ball
= ( 0 Red ball + other 3 balls ) + ( 1 Red ball + other 3 balls) + ( 2 Red ball + other 2 balls )
+ ( 3 Red ball + 1 other )
= no of ways to select 4 balls – no of select 4 red balls
\[ = {}^{12}{C_4} - {}^5{C_4}\]
\[\begin{array}{l}
= \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}} - 5\\
= 495 - 5\\
= 490
\end{array}\]
Hence the required number of ways is 490.
Note:
In this type of question students generally confuse at most and at least , more than and less than so take care of it during attempting these types of questions . Almost means max you get 3 red ball in a draw or minimum you can get zero
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
