On page 38, we claimed that since about a 1nfraction of n-bit numbers are prime, on average it is sufficient to draw O(n)random n -bit numbers before hitting a prime. We now justify this rigorously. Suppose a particular coin has a probability p of coming up heads. How many times must you toss it, on average, before it comes up heads? (Hint: Method 1: start by showing that the correct expression isi=1i(1-p)i-1p . Method 2: if E is the average number of coin tosses, show that E=1+(1-p)E).

Short Answer

Expert verified

1ptimes the coin must be tossed on average before it comes up heads.

Step by step solution

01

Explain the given information

Consider that,P is the Probability of a coin to show up heads. Let,X be the number of times the coin should be tossed before outcome is head. The following equation is used to find the average number of coin tossed:

E(X)=I=1i×P(i)

02

Calculate, how may time the coin must be tossed on average, before it comes up heads.

Consider that i-1times the coins were thrown , and the ith throwis a head. The probability of this event occurring is ,

pi=1-pi-1p

The number of time the coin must be tossed on average can be calculated as follows,

E=i=1i×PIE=i=1i1-pi-1p1-pE=i=1i1-pip

Simplify further,

localid="1659237715633" E-(1-p)E=p+i=11-pip=p+(1-p)pp=1

Find as follows,

E-1-pE=1E-E+pE=1pE=1E=1p

Therefore, 1ptimes the coin must be tossed on average before it comes up heads.

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