Question

A bag contains 2 red balls, 3 blue balls and 3 green balls find the probability to take

i) a Green Ball, ii) a Red Ball. When one ball is drawn random from the bag​

Answer 1

Answer:

$\large(i) \frac{3}{8}$

$(ii) \frac{1}{4}$

Step-by-step explanation:

$\bf \huge\underline \green{given} -$

• $2 \: red \: balls$
• $3 \: blue \: balls$
• $3 \: green \: balls$

$\huge \bf \underline \red{to \: find} -$

• $(i) \: Probability \: of \: a \: green \: ball$
• $(ii) \: Probability \: of \: a \: red \: ball$

$\huge\bf \underline\orange {Solution} -$

• $Total \: number \: of \: balls \: = 2 + 3 + 3$
• $Total number of balls = 8$
• $\large \: probability = \frac{number \: of \: favourable \: outcomes}{total \: number \: of \: outcomes}$
• (i) $probability \: of \: a \: green \: ball \: = \frac{3}{8}$
• (ii) $probability \: of \: a \: red \: ball \: = \cancel\frac{2}{8} = \frac{1}{4}$