The question is : 7 ballls are randomly withdrawn from an urn that contains 12re

Question

The question is : 7 ballls are randomly withdrawn from an urn that contains 12red, 16 blue and 18 green balls. find the probability that (a) 3 red, 3blue and 2 green balls are withdrawn (b) at least 2 red balls are withdrawn (c) all withdraw balls are the same colour (d) either exactly 3 red balls or exactly 3 blue balls arewithdrawn. I have tried the question, but don't know whether it corrector not, please provide the answer and same explaination if can.Thanks The question is : 7 ballls are randomly withdrawn from an urn that contains 12red, 16 blue and 18 green balls. find the probability that (a) 3 red, 3blue and 2 green balls are withdrawn (b) at least 2 red balls are withdrawn (c) all withdraw balls are the same colour (d) either exactly 3 red balls or exactly 3 blue balls arewithdrawn. I have tried the question, but don't know whether it corrector not, please provide the answer and same explaination if can.Thanks

Explanation / Answer

From an urn of 12 R , 16 B and 18 G , total number of ways ofwithdrawing 7 balls = # of ways of choosing 7 objects from(12+16+18)= 46 objects = 46C7 =
a) # of ways of selecting 3 R, 3 B and 2 G =( # of ways ofselecting 3 R out of 12 R balls)*( # of ways of selecting 3 B outof 16 B balls)*( # of ways of selecting 2 G out of 18 G balls) =12C3 * 16C3 * 18C2
P(# of ways of selecting 3 R, 3 B and 2 G) =( 12C3 * 16C3 *18C2 )/46C7 = 35.22%
b) # of ways of picking at least 2 red balls.
Other way to work it out is to see what's # of ways of getting0 Red and # of getting 1 Red
# of ways of getting 0 Red = # of ways of picking 7 balls from16 B and 18 G = 34 balls = 34C7 # of ways of getting 1 Red = (# of ways of picking 1 Red outof 12 R )* (# of ways of picking 6 balls from 16 B and 18 G = 34balls) = 12C1 *34C7
P( # of ways of picking at least 2 red balls) = 1-P(getting 0 R ) -P(getting 1 R) = 1-34C7/46C7- 12C1 *34C6/46C759.79%
C) # of ways of picking all balls of samecolor = # of ways of choosing all Red balls + # of ways ofchoosing all blue balls +# of ways of choosing all green balls =12C7 +16C7 +18C7
P(picking all balls of same color) =(12C7 +16C7 +18C7)/ 46C7  = 0.0823%
d) either exactly 3 R balls or exactly 3 Bballs are with drawn
# of ways of doing it is = # of drawing 3 R and4 other balls + # of ways of drawing 3 B and 4 other balls - # ofways of drawing 3 R , 3B and 1 G ball. (the last term is to avoid doublecounting)
= 12C3 * 34C4 + 16C3 * 30C4 - 12C3*16C3*18C1
P( either exactly 3 R balls or exactly 3 Bballs are with drawn) = (12C3 * 34C4 + 16C3 * 30C4 - 12C3*16C3*18C1 )/46C7 = 43.59%
Hope it helps .Feel free to ask for anyclarifications.


P(# of ways of selecting 3 R, 3 B and 2 G) =( 12C3 * 16C3 *18C2 )/46C7 = 35.22%
b) # of ways of picking at least 2 red balls.
Other way to work it out is to see what's # of ways of getting0 Red and # of getting 1 Red
# of ways of getting 0 Red = # of ways of picking 7 balls from16 B and 18 G = 34 balls = 34C7 # of ways of getting 1 Red = (# of ways of picking 1 Red outof 12 R )* (# of ways of picking 6 balls from 16 B and 18 G = 34balls) = 12C1 *34C7
P( # of ways of picking at least 2 red balls) = 1-P(getting 0 R ) -P(getting 1 R) = 1-34C7/46C7- 12C1 *34C6/46C759.79%
C) # of ways of picking all balls of samecolor = # of ways of choosing all Red balls + # of ways ofchoosing all blue balls +# of ways of choosing all green balls =12C7 +16C7 +18C7
P(picking all balls of same color) =(12C7 +16C7 +18C7)/ 46C7  = 0.0823%
d) either exactly 3 R balls or exactly 3 Bballs are with drawn
# of ways of doing it is = # of drawing 3 R and4 other balls + # of ways of drawing 3 B and 4 other balls - # ofways of drawing 3 R , 3B and 1 G ball. (the last term is to avoid doublecounting)
= 12C3 * 34C4 + 16C3 * 30C4 - 12C3*16C3*18C1
P( either exactly 3 R balls or exactly 3 Bballs are with drawn) = (12C3 * 34C4 + 16C3 * 30C4 - 12C3*16C3*18C1 )/46C7 = 43.59%
Hope it helps .Feel free to ask for anyclarifications.


P(picking all balls of same color) =(12C7 +16C7 +18C7)/ 46C7  = 0.0823%
d) either exactly 3 R balls or exactly 3 Bballs are with drawn
# of ways of doing it is = # of drawing 3 R and4 other balls + # of ways of drawing 3 B and 4 other balls - # ofways of drawing 3 R , 3B and 1 G ball. (the last term is to avoid doublecounting)
= 12C3 * 34C4 + 16C3 * 30C4 - 12C3*16C3*18C1
P( either exactly 3 R balls or exactly 3 Bballs are with drawn) = (12C3 * 34C4 + 16C3 * 30C4 - 12C3*16C3*18C1 )/46C7 = 43.59%
Hope it helps .Feel free to ask for anyclarifications.

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