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Chapter 4: Q.4.15 (page 170)

Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is

121+(q-p)n.whereq=1-p.Do this by proving and then utilizing the identity

∑i=0[n/2]n2ip2iqn-2i=12(p+q)n+(q-p)n

Short Answer

Expert verified

In the given information the answer is =121+(q-p)nisproved

Step by step solution

01

Given Information

We know that ,

p+qn=∑i=0nCknpkqn+1(1)

q-pn(q-p)n=∑i=0nCkn(-p)kqn-k(2)

02

Calculation

When we add these two expressions ,

⇒p+qn+q-pn=2∑i=0n2C2ip2iqn-2i

because when kis odd -pkis negative so (1) and (2) will cancel .

⇒∑i=0n2C2ip2iqn-2i=12(p+q)n+(q-p)n

⇒pEvenheads=12(p+1-p)n+(q-p)n

=121+(q-p)n hence proved

03

Final Answer

The answer is=121+(q-p)nis proved121+(q-p)n

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Most popular questions from this chapter

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