Question

IF A BAG CONTAINS 3 RED BALLS AND 6 BLUE BALLS .WHAT IS THE PROBABILITY THAT
1)BLUE BALL CAME OUT
2)RED BALLCAME OUT​

Answer 1

Step-by-step explanation:

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Assuming every single ball has an equal chance of being drawn, then if B is the probability of drawing a blue ball and R is the probability of drawing a red ball:

B + R = 1,

since either a red or blue ball will be drawn. Since each ball has an equal chance of being drawn:

16 p = 1, or

p = 1/16

where p is the chance to draw a single ball. Since there are 10 red balls, there are ten independent and different ways to draw a red ball, which means that

R = 10 p = 10/16 = 5/8.

You can either apply the first equation to find that

B = 1 - R = 3/8

or state directly that

B = 6 p = 6/16 = 3/8

since there are six independent and different ways to draw a blue ball from the bag.

Answer 2

Answer:

$probability \: of \: blue \: balls \: = \frac{6}{9} = \frac{2}{3} \\ probability \: of \:red \: balls \: \frac{3}{9} = \frac{1}{3}$

Step-by-step explanation:

1}PROBABILITY OF GETTING BLUE BALLS

$= \frac{no \: of \: blue \: balls}{total \: no \: of \: balls} \\ = \frac{6}{9} \\ = \frac{2}{3}$

2}PROBABILITY OF GETTING RED BALLS

$\frac{no \: of \: red \:balls}{total \: no \: of \: balls} \\ = \frac{3}{9} = \frac{1}{3}$