## Amazon Interview Question for SDE1s

• -6

Country: India
Interview Type: Written Test

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3
of 3 vote

pA = 6 / 7;
pB = 4 / 5;
pC = missing in the statement.

We can pick (A,B), (A,C), (B,C), so answer = (pA*pB + pA*pC + pB * pC) / 3

Comment hidden because of low score. Click to expand.
-1
of 1 vote

instead of multiplying the whole term by 1/3 you should multiply it with 2/3 because everytime you are selecting 2 out of 3 persons.

Comment hidden because of low score. Click to expand.
2

There are C(3,2) = 3 ways to pick 2 out of 3 people, disregarding order. Then, there's a 1/3 chance of picking one of those 3 cases.

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0
of 0 vote

P(a intersection b)=p(a)+p(b)-p(a union b)
Am i correct regarding this case?

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0
of 0 vote

C shot details??

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0

Right :) C's details?

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0

C details : assume that C hits 4 times out of 9

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0
of 0 vote

if we consider a c probability 3 out of 4 times then
PA =6/7
PB = 4/5
PC = 3/4
total probability is PA * PB * PC(not hitting) + PA*PB(not hitting)*PC + PA(not hitting) * PB*PC
sp probability is = 6/7 * 4/5 * 1/4 + 6/7*1/5*3/4 + 1/7*4/5*3/4 =27/70

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0

Can you elaborate?
!A, !B or !C is the probability, that they do not hit the target, not that they do not shoot, while the probability that AC, BC or BC is going to happen is 1/3 each.

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0
of 0 vote

there are three ways to choose three ppl out of three: AB, AC, and BC.
P(hitting target twice) = P(first chosen person hit) P(second chosen person hit) = P(A hit)P(B hit) + P(A hit)P(C hit) + P(B hit)P(C hit)

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0
of 0 vote

this seems a incomplete problem and to solve it, more information is needed.
For instance, how many times can the two persons try? 2, or more than 2?

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0
of 0 vote

if we assume that C hits 3 out of 4 times then :-
ans = 1/3( 6/7*4/5 + 4/5*3/4 + 3/4*6/7)

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0

why do we need to divide by 3??

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0
of 0 vote

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