Question

A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. what is the probability that the balls contain no yellow ball?

Answer 1

Probability of no yellow ball=3+4/5+3+4=7/12

Answer 2

Answer:

$Probability \:that \:the \:balls \: contains \\no \: yellow \:ball = p( \overline Y) = \frac{ 21}{22}$

Step-by-step explanation:

$Number \: of \: balls \: in \:a \: bag\:are :$

$Red = 3,\\Yellow = 5 , \\Green = 4$

$Total\: balls \: in \: a \: ball = 3+5+4 = 12$

$Probability \:that \:the \:balls \: contains \\no \: yellow \:ball = p( \overline Y) = 1 - \frac{^{5}C_{3}}{^{12}C_{3}}$

$= 1 - \frac{10}{220}$

$= 1 - \frac{1}{22}$

$= \frac{22 - 1}{22}$

$= \frac{ 21}{22}$

Therefore.,

$Probability \:that \:the \:balls \: contains \\no \: yellow \:ball = p( \overline Y) = \frac{ 21}{22}$

•••♪