Question

Find the value of above variables, m and n
with explanation. ​

Answer 1

$\rule{300}{3}$

GIVEN :-

⁂ $\textsf{A bag contains 'm' white balls.}$

⁂ $\textsf{It contains 21 red balls.}$

⁂ $\textsf{It also contains 'n' blue balls.}$

$\longrightarrow\textsf{ A ball is drawn at random from the bag.}$

$\bigstar\ \textsf{Probability of getting a blue ball is }\displaystyle{\sf{\frac{1}{4}. }}$

$\bigstar\ \textsf{Probability of getting a blue ball is }\displaystyle{\sf{\frac{7}{20}. }}$

TO FIND :-

$\textsf{The value of 'm' and 'n'.}$

SOLUTION :-

$\textsf{Total number of balls = 21 +m+n}$

$\displaystyle{\sf{Probability\ of\ getting\ a\ red\ ball=\frac{Number\ of\ red\ balls}{Total\ number\ of\ blue\ balls} }}$

$\displaystyle{\sf{\implies \frac{7}{20}=\frac{21}{21+m+n} }}\\\\\\\displaystyle{\sf{\implies \frac{1}{20}=\frac{3}{21+m+n} }}\\\\\\\displaystyle{\sf{\implies 21+m+n=60}}\\\\\displaystyle{\sf{\implies m+n=39}}\\\\\displaystyle{\sf{\implies m=39-n}}\:\:\:\:\:\:\:\: \cdots(i)$

$\\\\\displaystyle{\sf{Probability\ of\ getting\ a\ blue\ ball=\frac{Number\ of\ blue\ balls}{Total\ number\ of\ blue\ balls} }}\\\\\displaystyle{\implies \sf{\frac{1}{4}=\frac{n}{21+m+n} }}\\\\\displaystyle{\sf{\implies 21+m+n=4n}}\\\\\displaystyle{\sf{\implies 21+m-3n=0}}\\\\\displaystyle{\sf{\implies m=3n-21}}\\\\\displaystyle{\sf{\implies 39-n=3n-21}}\:\:\:\:\:\:\:\: \cdots\sf{[from\ (i)]}\\\\\displaystyle{\sf{\implies 60=4n}}\\\\\underline{\boxed{\displaystyle{\sf{\implies n=15}}}}$

$\sf{Now,\ from\ (i),\ m=39-n.}\\\\\textsf{Putting the value of n = 4 in (i) :}\\\\\displaystyle{\sf{m=39-4}}\\\\\underline{\boxed{\displaystyle{\sf{\implies m=35 }}}}$

$\rule{300}{3}$

Answer 2

Answer:

above answer is correct mark it as brainliest one.