Question

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag What
is the probability that the ball drawn is (i) red ? (ii) not red?
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Answer 1

the answer is in above pic hope it helps

Answer 2

Answer :-

• The probability P(E) that the ball drawn is Red is 3/8
• The probability P(E) that the ball drawn is Not Red is 5/8

Given :-

• Number of Red Balls in Bag = 3
• Number of Black Balls in Bag = 5

To Find :-

• The probability P(E) that the ball drawn is Red ?
• The probability P(E) that the ball drawn is Not Red ?

Explanation :-

As per above Data,

$\sf \implies Total \: Number \: of \: Balls \: in \: Bag = 3 + 5 = \red{8 \: Balls}$

AS we know,

$\green { \boxed { \odot \: \small \bf { Probability \: of \: an \: Event \: P(E)} = \small \text { \bf { \dfrac{Total \: No. \: of \: Favorable \: Outcome}{Total \: No. \: of \: Outcome}} } }}$

Now, Substituting the values,

$\blue { \sf \bigstar \: \text{Probability of Getting a } \red{Red \: Ball} }$

$\sf : \: \longrightarrow P(E) \: of \: getting \: a \: red \: ball = \dfrac{Total \: No. \: of \: Favorable \: Outtcome}{Total \: No. \: of \: Outtcome}$

$\red { \underline { \boxed { \sf : \: \longrightarrow P(of \: getting \: a \: red \: ball) = \dfrac{3}{8} }}}$

$\blue { \sf \bigstar \: \text{Probability of Getting a not a red ball i.e. }} \sf Black \: Ball$

$\sf : \: \longrightarrow P(E) \: of \: getting \: a \: not \: red \: ball \: i.e. \: black \: ball = \dfrac{Total \: No. \: of \: Favorable \: Outcome}{Total \: No. \: of \: Outcome}$

$\underline { \boxed { \sf : \: \longrightarrow P(of \: getting \: a \: not \: a \: red \: ball \: i.e. \: black \: ball) = \dfrac{5}{8} }}$

Hence, the Probability of getting a Red Ball and Not getting a Red Ball are 3/8 and 5/8 respectively.

@Agamsain