Help with this please
1 answer:
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Answer:
0.32
Step-by-step explanation:
Let A be the event that the transferred ball was blue.
B be the event that the red ball is selected from box 2
C be the event that the transferred ball was red
Then:
P(A) be the probability that the transferred ball was blue, which is 3/8
P(C) be the probability that the transferred ball was red, which is 5/8
P(B|A) is the probability that selected ball is red, given that the transferred ball was blue, which is 4/7
P(B|C) is the probability that selected ball is red, given that the transferred ball was red, which is 5/7
P(B) is the probability that the selected ball is red, which there are 2 scenarios, given that the transferred ball is blue and red
P(B) = P(B|A)P(A) + P(B|C)P(C)
P(A|B) is the probability that the transferred ball was blue, given that a red ball is selected from Box 2, which can be solved using Bayes theorem
Answer:
y=9x+20
Step-by-step explanation:
5x-12y-17=0 (plus 17 from both sides)
5x-12y=17 (plus 12y from both sides)
5x=12y+17 (divide 5 from everything)
5x-12y-17=0 (plus 17 from both sides)
5x-12y=17 (minus 5x from both sides)
-12y=-5x+17 (divide everything by -12)
1*9(x+2)-(y+1)+3=0 (distribute 9 to x and to 2) (distribute -1 to y and to 1)
1*9x+18-y-1+3=0 (times 9x by 1)
9x+18-y-1+3=0 (combine like terms)
9x-y+20=0 (minus 20 from both sides)
9x-y=-20 (plus y to both sides)
9x=y-20 (divide 9 from both sides)
1*9(x+2)-(y+1)+3=0 (distribute 9 to x and to 2) (distribute -1 to y and to 1)
1*9x+18-y-1+3=0 (times 9x by 1)
9x+18-y-1+3=0 (combine like terms)
9x-y+20=0 (minus 20 from both sides)
9x-y=-20 (minus 9x from both sides)
-y=-9x-20 (divide -1 from everything)
y=9x+20