Question

a bag contains 5 red balls 6 white balls 7 green balls 8 black balls. one ball is drawn at random for the bag .find the probability that the ball drawn is
1)white
2)black or red
3)not white
4)neither white nor black ​

Answer 1

Answer:

i)3/13

ii)½

iii)10/13

iv)6/13

Step-by-step explanation:

Total number of balls=5+6+7+8=26

P(white balls)=6/26=3/13

P(black or red)=13/26=½

P(not white)=20/26=10/13

P(neither white nor black)=12/26=6/13

Answer 2

1) The probability that the ball drawn is white = $\dfrac{3}{13}$

2) The probability that the ball drawn is black or red =  $\dfrac{1}{2}$

3) The probability that the ball drawn is not white =  $\dfrac{10}{13}$

4)  The probability that the ball drawn is neither white nor black= $\dfrac{6}{13}$

Explanation:

Given :  A bag contains 5 red balls 6 white balls 7 green balls 8 black balls.

Total balls =  26

If one ball is drawn at random for the bag .

1) The probability that the ball drawn is white = $\dfrac{\text{No. of white balls}}{T}$

$=\dfrac{6}{26}=\dfrac{3}{13}$

The probability that the ball drawn is white = $\dfrac{3}{13}$

2) The probability that the ball drawn is black or red = $\dfrac{\text{Total No. of black and red balls}}{T}$

$=\dfrac{8+5}{26}=\dfrac{13}{26}=\dfrac{1}{2}$

The probability that the ball drawn is black or red =  $\dfrac{1}{2}$

3) The probability that the ball drawn is not white:=

$1-P(white)= 1-\dfrac{3}{13}=\dfrac{10}{13}$

The probability that the ball drawn is not white =  $\dfrac{10}{13}$

4) The probability that the ball drawn is neither white nor black ​:-

$\dfrac{\text{No. of balls other than white and black}}{T}$

$=\dfrac{26-6-8}{26}=\dfrac{12}{26}=\dfrac{6}{13}$

The probability that the ball drawn is neither white nor black= $\dfrac{6}{13}$

# Learn more :

A bag contains 5 red balls,3 black balls and 4 white balls a ball is drawn at random.what is the probability that the ball drawn is a neither white nor black

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