Question

A box contains 5 chips, numbered 1, 2, 3, 4, and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required?

Answer 1

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Answer 2

Answer:

1/5

Step-by-step explanation:

Total number of chips= total possibility=5

P(sum >4, in 3 draws), without replacement=

[P(1) and P(2) and P(3 or 4 or 5)] =1/20

Or

[P(1) and P(3 ) and P(2 or 4 or 5)= 1/20

Or

[P(2) and P(1) and P(3 or 4 or 5)]=1/20

Or

[P(3 ) and P(1) and P(2 or 4 or 5)]=1/20

Therefore , P(sum>4 ,in 3 draws) ,without replacement = (1/20)+(1/20)+(1/20)+(1/20) =

4*(1/20)=1/5.