Question

There are 18 beads in a box. Some of them are white and the remaining are black. The probability of drawing a black from it is 1/3. Then
a. How many black beads are there in the box?

b. How many white beads are there in the box?

c. How many white beads should be added to it, so that the probability of drawing a black bead become 1/4?

Answer 1

$Total beads = 18 \\ = </p><p>P(Black)= \frac{1}{3} = \frac{6}{18} \\ no \: of \: black \: balls = 6 \\ no \: of \: white\: balls = 18 - 6 = 12 \\ - - - - - - - - - - - - - - \\ let \: x \: white \: balls \: to \: be \: added \: to \: make \: \\ prob \: of \: black = \frac{1}{4} \\ prob \: of \: white = 1 - \frac{1}{4} = \frac{3}{4} \\ \frac{12 + x}{18 + x} = \frac{3}{4} \\ 48 + 4x = 54 + 3x \\ x = 6 \\ hence \: 6 \: white \: ball \: should \: be \: added$

Mark my ANS BRILLIANT

Answer 2

Answer:

Step-by-step explanation:

a)6/18=1/4

Therefore their are 6 black beads in the box

b)12+6=18

Therefore their are 12 white beads in the box

Answer 3

Answer:

Step-by-step explanation:

a)6/18=1/4

Therefore their are 6 black beads in the box

b)12+6=18

Therefore their are 12 white beads in the box