A bag comtains 15 ball of which X are blue and the remaining are red . If yhe number of balld are increased by 5, the probability of drawing the red ball doubles, find (1) P(red ball) (2) P (Blue ball) (3) P( Blue ball it of 5 extra ted balls are actually added
Answers
Total number of balls in the bag = N = 15.
Number of blue balls = X.
Number of red balls = 15 - X.
Probability of drawing a red ball
= favourable outcomes/total number of outcomes
= (15 - X)/15
Increase the number of Red balls by 5.
So number of red balls = 15 - X + 5 = 20 - X
Probability of drawing a red ball
P(Red ball) = (20 - X)/(15+5) = (20 - X)/20
Given (20 - X)/20 = 2 * (15 - X)/15
3(20 -X) = 8 (15 - X)
5 X = 60
X = 12
So there are 12 blue balls and 3 red balls. A total of 15 balls.
1) P(red ball) = initial = 3/15 = 1/5.
after adding 5 red balls, 2/5
2) P(blue ball) = initial = 12/15 = 4/5.
after adding 5 red balls, 12/20 = 3/5
There is data missing in this question or there is a word missing. It should read "If the number of RED balls are increased by 5, ..."
Total number of balls in the bag = N = 15.
Number of blue balls = X.
Number of red balls = 15 - X.
Probability of drawing a red ball
= favourable outcomes/total number of outcomes
= (15 - X)/15
Increase the number of Red balls by 5.
So number of red balls = 15 - X + 5 = 20 - X
Probability of drawing a red ball
P(Red ball) = (20 - X)/(15+5) = (20 - X)/20
Given (20 - X)/20 = 2 * (15 - X)/15
3(20 -X) = 8 (15 - X)
5 X = 60
X = 12
So there are 12 blue balls and 3 red balls. A total of 15 balls.
1) P(red ball) = initial = 3/15 = 1/5.
after adding 5 red balls, 2/5
2) P(blue ball) = initial = 12/15 = 4/5.
after adding 5 red balls, 12/20 = 3/5