Question

An urn contains 3 red, 4 green and 5 white balls. A ball is selected
randomly. What is the probability of selecting (i) red ball, (ii) non red
ball and (iii) red or green ball?​

Answer 1

Solution

Givens:

3 red

4 green

5 whites ball

Answer:

3/12 probability of selection red

1/12 probability of selecting not red

7/12 probability of selection red or green

Step-by-step explanation:

I hope this helped!

Answer 2

SolutioN :

• Number of Red balls = 3
• Number of Green Balls = 4
• Number of white balls = 5
• Total balls = 5 + 4 + 3 = 12

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1) Selecting a Red ball

$\longrightarrow \sf{Probablility \: = \: \dfrac{No. \: of \: Red \: balls}{Total \: no. \: of \: balls}} \\ \\ \longrightarrow \sf{Probability \: = \: \dfrac{3}{12}} \\ \\ \longrightarrow \sf{Probability\: = \: \dfrac{1}{4}}$

• Probability of getting a Red ball is 1/4

_____________________

2) Non-Red Ball

⇒Number of non Red balls = 12-3 = 9

$\longrightarrow \sf{Probability \: = \: \dfrac{No. \: of \: Non-red \: balls}{Total \: no. \: of \: balls}} \\ \\ \longrightarrow \sf{Probabiliy \: = \: \dfrac{9}{12}} \\ \\ \longrightarrow \sf{Probability \: = \: \dfrac{3}{4}}$

• Probability of getting a Non-Red ball is 3/4

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3) Red or Green ball

⇒ No. of Red or Green ball = 3 + 4 = 7

And if A green or a Red ball :

$\longrightarrow \sf{Probability \: = \: \dfrac{No. \: of \: green \: or \: Red \: ball}{Total \: no. \: of \: balls}} \\ \\ \longrightarrow \sf{Probability \: = \: \dfrac{7}{12}} \\ \\ \longrightarrow \sf{Probability \: = \: \dfrac{7}{12}}$

• Probability of getting a green ball is 7/12