Question

There are 24 balls in a pot. If 3 of them are Red,5 of them are Blue and remaining are Green then,

what is the probability of picking out (i) a Blue ball,(ii) a Red ball and (iii) a Green ball?​

Answer 1

Before, finding the answer. Let's find out on how we can find the answer.

• Let's first know that,

Probability = $\sf \dfrac{no \: of \: outcomes }{total \: no \: of \: outcomes}$

• So, we can write by using the above formula.
• And for green, we have to add red and blue. Then, subtract from total number.

______________________

Given :

• Total Balls = 24 balls
• Red = 3 balls
• Blue = 5 balls
• Green = Remaining Balls

To find :

⇒ Probability of picking out

• a Blue ball
• a Red ball
• a Green ball

Solution :

Green Ball = Total Balls - (Blue ball + Red ball)

                  = 24 - (5 + 3)

                  = 24 - 8

                  = 16 balls

Probability of Blue Ball  = $\sf \dfrac{no \: of \: outcomes }{total \: no \: of \: outcomes}$

                                       = $\sf \dfrac{Blue \: Balls }{Total \: Balls }$

                                       = $\sf \dfrac{5}{24 }$

Probability of Red Ball = $\sf \dfrac{no \: of \: outcomes }{total \: no \: of \: outcomes}$

                                     =  $\sf \dfrac{Red \: Balls }{Total \: Balls }$

                                     =  $\sf \dfrac{3}{24 }$

Probability of Green Ball = $\sf \dfrac{no \: of \: outcomes }{total \: no \: of \: outcomes}$

                                         = $\sf \dfrac{Green \: Balls }{Total \: Balls }$

                                         = $\sf \dfrac{16}{24 }$