One ticket is drawn at random from each of the two boxes.(A) 1 2 3 4 5

(B) 1 2 3 4 5 6

Find the chance that

(a) one of the number is 2 and the other is 5

(b) sum of the numbers is 7

(c) one number is bigger than twice the other

Answers

Answer 1
Answer:

Answer:

a.1/15

b.1/6

c.1/3

Step-by-step explanation:

number of outcomes from box 1=5

number of outcomes from box 2=6

therefore total number of outcomes=6×5

=30

a. number of times receiving a ticket 2 and the other 5 =2

therefore probability= 2/30

=1/15

b.number of combinations for 7= 5 (1+6, 2+5,5+2,4+3,3+4.)

therefore probability= 5/30

=1/6

c.number of times one number was bigger than twice the other=10 (1+3,3+1,1+4,4+1,1+5,5+1,1+6,2+5,5+2,2+6)

therefore probability= 10/30

=1/3


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Answers

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

First you multiply both sides of the equation by 5 to get rid of the fraction, it becomes Y + 30 = -50.

To get rid of the 30 you subtract 30 from each side and that makes it Y = -80.

Your answer does not want the Y=, so your answer is -80.

If this helps, please mark it Brainliest :)