pg612 Abbys parked in a mile marker on and East West Country

pg612

Abby\'s parked in a mile marker on and East- West Country Rd. She decides to toss a fair coin10 times, each time driving 1 mile east if it lands heads up and 1 mile west if it lands tales up. The term \"random walk\" applies to this process, even though Abby drives rather than walks. It is a simplified model of Brownian motion.

a). what is the probability that abbys \"walk\" will end 6 miles east of the start?

b). what is the probability that abbys \"walk\" will end at least 2 miles east of the start?

Solution

Given

Abby\'s parked in a mile marker on and East- West Country Rd. She decides to toss a fair coin10 times, each time driving 1 mile east if it lands heads up and 1 mile west if it lands tales up. The term \"random walk\" applies to this process, even though Abby drives rather than walks. It is a simplified model of Brownian motion.

A) we have to find the probability that abbys \"walk\" will end 6 miles east of the start

let X denote no. of miles abbys walks in east from start

since abbys tosses 10 times so he walks 10 times(one mile each time) sometimes to east and sometimes to the west .

since here resulting dstance from start is 6 so he must be get 8 times heads and 2 times tail so

we have to find P(X=8)=?

where X follows bionomial distribution with p=0.5 and n=10 (because experiment depends on tossing of fair coin and coin is tossed 10 times)

now

P(X=8)=(10!/8!*2!) (0.5)⁸ (0.5)²

        =45*(0.5)¹⁰

        =45*0.0009765625

        =0.0439453125      answer

B) we have to find the probability that abbys \"walk\" will end at least 2 miles east of the start

for this on tossing at least 6 heads occurs

so we have to find

P(X>6)=?

now

P(X>6)=P(X=6) +P(X=7) +P(X=8) +P(X=9) +P(X=10)

         =(10!/6!4!)(0.5)¹⁰ + (10!/7!3!)(0.5)¹⁰ +(10!/8!2!)(0.5)¹⁰ + (10!/9!1!)(0.5)¹⁰ +(10!/10!0!)(0.5)¹⁰

        =(0.5)¹⁰ (210 +120 +45 +10 +1)

    = (0.5)¹⁰(386)

   =0.3770   answer