Abag contains 144 ping-pong balls. more than half of the balls are painted orange and the rest are painted blue. two balls are drawn at random without replacement. the probability of drawing two balls of the same color is the same as the probability of drawing two balls of different colors. how many orange balls are in the bag?

Let us say that:

x = number of orange balls

144 –x = number of blue balls

From the given statement, we can create the equations:

probability of drawing two orange balls = probability of drawing two different colored ball

(x / 144) [(x – 1) / 143] = (x / 144) [(144 –x) / 143] + [(144 –x) / 144] (x / 143)

Divide everything by x, and multiply everything by 144*143:

(x – 1) = (144 – x) + (144 –x)

x – 1 = 288 – 2x

3x = 289

x = 96.33 ~ 96

144 – x = 48

So there are about 96 orange balls

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

its c or d   take a lucky guess my dude