A bag contains 5 white and 8 red balls. two drawing of 3 balls are made such that (a) balls are replaced before the second trial (b) the balls are not replaced before the second trial. find the probability that the first drawing will give 3 while and the second 3 red with in each case. plz answer, plz answer
Answers
Step-by-step explanation:
Total number of balls in the bag = 13.
3 balls can be drawn in C(13,3) ways.
(a) Given that there are 5 white balls, 3 balls are drawn.
= C(5,3) ways.
(b) Given that there are 8 red balls, 3 balls are drawn.
= C(8,3) ways.
(i) Balls are replaced:
Probability of drawing 3 white balls in the first trial:
= C(5,3)/C(13,3)
= 10/286
= 5/143
Probability of drawing 3 red balls in the second trial:
= C(8,3)/C(13,3)
= 56/286
= 28/143
∴ Required Probability = (5/143) * (28/143)
= 140/20449.
(ii) Balls are not replaced:
Probability of drawing 3 white balls in the first trial:
= C(5,3)/C(13,3)
= 10/286
= 5/143
When the 3 white balls are drawn without replacement, the bag contains only 2 white balls and 8 red balls.
In the second trial, 3 balls can be drawn from 10 balls(8 + 2) in C(10,3) ways and 3 red balls can be drawn from 8 balls in C(8,3) ways.
Probability of drawing 3 red balls in the second trial:
= C(8,3)/C(10,3)
= 56/120
= 7/15
∴ Required Probability = (5/143) * (7/15)
= 35/2145
= 7/429
Hope it helps!
First three balls are white, with no replacement (5/13)(4/12)(3/11)= 60/1716
Now the second draw are red (8/10)((7/9)(6/8)= 336/720.
Multiply these, for this is a joint probability
(60/1716) * (336/720)
Cancel 60 and 720
(1/1716) * (336/12)
Work on second term
(1/1716)*28
Divide by 4 numerator and denominator
(7/429)