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Here the question is about "dependency" not hazard ..So we see by Bernstein's conditions..So for RAW dependency , we need to check that write of an earlier instruction is succeeded by some subsequent instruction and following this we have 4 dependencies

a) I1 - I2

b) I1 - I3

c) I2 - I3

d) I2 - I4..

So in all we have 4  RAW dependencies here..So correct answer..Remember that hazard is not same as dependency..Hazard finding needs other factors also like number of stages etc..

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