An experiment consists of observing the contents of a 16-bit register. We assume

Question

An experiment consists of observing the contents of a 16-bit register. We assume that all 65536 byte values are equally likely to be observed. Let A denote the event that the Least Significant Bit (LSB) is a ZERO. What is P(A)? Let B denote the event that the register contains 5 ONEs and 11 ZEROs. What is P(B)? What is the probability of at least one of A or B occurs, i e. what is P (A B)? What is the probability that exactly one of A and B occur, i e. what is P (A B)? What is the probability of at least one of A or B does not occur?

Explanation / Answer

a) as right most bit is 0, rest 15 digits have 2 choices (0 or 1)

hence total number of outome for event A =215

and total outcome =216

therefore P(A) =215/216 =(1/2)=0.5

b)from multinomial , number of ways to divide 16 numbers in two types of 5 and 11 =16!/(5!*11!) =4368

hence probability P(B) =4368/216 =0.06665

c)the common number of ways b/w A and B when LSB =0. therefore we have to arrange 15 digits in 5 one and 10 zero

number of ways =15!/(10!*5!) =3003

hence probabilty P(AnB) =3003/216 = 0.0458

P(AUB) =P(A)+P(B)-P(AnB) =0.5+0.06665-0.0458 =0.5208

d) probabilty that exactly one will occur =P(A)+P(B)-2*P(AnB)=0.5+0.06665-2*0.0458 =.475

e)probabilty of at least one A or B does not occur =1-P(both will occur) =1-0.5208 =0.4792

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