Two balls are drawn in succession out of a box containing 2 red and 5 white balls. Find the probability that at least one ball w
as red , given that the first ball was:
(A) replaced before the second draw.
(B) not replaced before the second draw.
1 answer:
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➷ A) (2/7 x 2/7) + (2/7 x 5/7) + (5/7 x 2/7) =
4/49 + 10/49 + 10/49 = 24/49 <== this is the answer
B) (2/7 x 1/6) + (2/7 x 5/6) + (5/7 x 2/6) -
2/42 + 10/42 + 10/42 = 22/42 <== this is the answer
The second answer could be simplified to 11/21 (only write this answer if it wants the simplified version, otherwise write the one above)
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Answer:
15
Step-by-step explanation:
H - 7
When H = 22, replace H with 22 in the equation.
22 - 7 = 15
I believe the next one would be 13.2 because if you add 8.4 to -6.4 it gets 4.8 so you just do the same thing
So you do the recipical and multipy 7 by 2 and your fraction is 14/24 and simplify which you get 7/12
Answer:
5 times bigger
Step-by-step explanation:
(6-4) = 2
(6-4) x 5 = 10
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