Question

There are 8 red and 3 black balls in a bag. What is the probability of drawing a red ball​

Answer 1

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TOTAL NO OF BALLS = 8+3=11

TOTAL NO OF FAVOURABLE OUTCOMES=8 (SINCE THERE ARE 8 RED BALLS)

P(drawing a red ball) = 8/11

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Answer 2

Given Question :-

• There are 8 red and 3 black balls in a bag. What is the probability of drawing a red ball?

Answer

Given :-

• Bag having 8 red and 3 black balls.

To find :

• Probability of drawing a red ball.

Concept used :-

Assume n(R) as number of red balls and n(B) as number of black balls. To find the number of balls, take the sum of number of red balls and number of black balls and denote it by n(S). Now, to find the probability of drawing a red ball, take the ratio of n(R) to n(S).

Let's do it now!!

Solution :-

Let us assume n(R) and n(B) denotes the number of red balls and number of black balls in bag. Also assume that total number of balls is denoted by n(S).

Now, it is given that bag contains 8 red and 3 black balls.

$\tt\implies \:n(R) \: 8 \: and \: n(B) = 3$

So, total number of balls in the bag will be the sum of red and black balls, therefore,

$\tt\implies \:n(S) = n(R) + n(B) \\ \tt\implies \:n(S) = \: 8 \: + \: 3 \: \: \: \: \: \: \:$

$\tt\implies \:n(S) = 11$

Now, we have to find the probability of drawing a red ball from bag. So, required probability will be the ratio of number of red balls to total number of balls in bag.

$\tt\implies \:P(R) = probability \: of \: drawing \: red \: ball = \dfrac{n(R)}{n(S)}$

$\tt\implies \:P(R) = \dfrac{8}{11}$