A walker decides to move up or down using unfair coin. The coin tells the walker

Question

A walker decides to move up or down using unfair coin. The coin tells the walker to move up 1 step with probability 0.6 and down 1 step with probability 0.4. The walker tosses the coin prior each step. The walk stops when the walker reaches 5 step line or -2 step line.

Answer the following questions:

Question1(C) What is the probability that the walker will stop after the step 1?

Question2(C) What is the probability that the walker will stop after the step 2?

Question3(C) What is the probability that the walker will stop after the step 3?

Question4(C) What is the probability that the walker will stop after the step 4?

Question5(C) What is the probability that the walker will stop after the step 5?

Question6(B) What is the probability that the walker will stop after the step 6?

Question7(B) What is the probability that the walker will stop after the step 7?

Question8(B) What is the probability that the walker will stop after the step 8?

Question9(B) What is the probability that the walker will stop after the step 9?

Question10(B) What is the probability that the walker will stop after the step 10?

Question11(A) What is the probability that the walker will stop after the step 11?

Question 12(A) Using information from the previous 11 questions determine which stopping step is the most probable and what is the expected number of steps needed to stop?

Explanation / Answer

let us assume if head appears then the walker goes one step up and if tail appears then the walker goes one step down with P[Head]=0.6 P[tail]=0.4

so the steps of the walker are independent of each other.

the walk stops when the walker reaches 5 steps line or -2 step line

let X denotes the number of step downs the walker takes in n steps.

then X~Bin(n,0.4) so P[X=x]=nCx0.4x0.6n-x x=0,1,2,3,..........,n

question 1C
the probability that the walker will stop after the step 1 is 0
because after step 1 either he will go one step up or 1 step down.
there is no way that walker can reach at 5 steps up or 2 steps down.

question 2C
the walker will stop after step 2 means in 2 steps he has either reached -2 step line or 5 step line.
now 5 step line is impossible in 2 steps. so the only possible case is both the times tail have appeared.
so the required probability is P[X=2] where X~Bin(2,0.4)

=2C20.42=0.16 [answer]

question 3C
in 3 steps, 4 cases may occur
all 3 are heads : walker goes 3 steps line
2 heads 1 tail: walker goes +2-1=1 step line
1 head 2 tails: walker goes +1-2=-1 step line
3 tails: walker goes -3 steps line
so there is no possibiliy to reach 5 steps line or -2 steps line. so the  probability that the walker will stop after the step 3 is 0 [answer]

question 4C

here 5 cases may happen
4 heads: walker goes 4 steps line
3 heads 1 tail: walker goes +3-1=2 steps line
2 heads 2 tails: walker goes +2-2=0 steps line
1 head 3 tails: walker goes +1-3=-2 steps line
4 tails: walker goes -4 steps line

so only chance to stop the walker is if 1 head and 3 tails appear.

so required probability is P[X=3] where X~Bin(4,0.4)

=4C30.430.61=0.1536

5C
here 6 cases may happen
5 heads: 5 steps line
4 heads 1 tail : 3 steps line

3 heads 2 tail: 1 step line
2 heads 3 tails: -1 step line
1 head 4 tails: -3 step line
5 tails: -5 step line
so probability of stopping is P[5 heads appear]=P[X=0] X~Bin(5,0.4)

=5C00.65=0.07776

6B
it is impossible to go to 5 steps line with 6 steps . he may reach at -2 steps line if there are 4 tails and 2 heads

so probability of stopping=P[X=4] where X~Bin(6,0.4)

=6C40.440.62=0.13824 [answer]

7B
it is impossible to reach at -2 steps line at 7 steps. he may reach at 5 steps line if 6 heads and 1 tails appears
so required probability is 7C10.4*0.66=0.1306368

8B

it is impossible to go to 5 steps line with 8 steps . he may reach at -2 steps line if there are 5 tails and 3 heads

so probability of stopping=8C60.450.63=0.06193152

9B

it is impossible to go to -2 steps line with 9 steps . he may reach at 5 steps line if there are 2 tails and 7 heads

so probability of stopping=9C20.420.67=0.161243136

10B

it is impossible to go to 5 steps line with 10 steps . he may reach at -2 steps line if there are 6 tails and 4 heads

so probability of stopping=10C60.460.64=0.111476736

11B

it is impossible to go to-2 steps line with 11 steps . he may reach at 5 steps line if there are 3 tails and 8 heads

so probability of stopping=11C30.430.68=0.177367449

12A
so from above 11 questions we get

stopping step probability

1 0

2 0.16

3 0

4 0.1536

5 0.07776

6 0.13824

7 0.1306368

8 0.06193152

9 0.161243136

10 0.111476736

11 0.177367449

so most probable stopping step is 11 because it has the maximum probability of 0.177367449

so expected number steps needed to stop:

1*0+2*0.16+3*0+4*0.1536+5*0.07776+6*0.13824+7*0.1306368+8*0.06193152+9*0.161243136+10*0.111476736+11*0.177367449=8.079547283 [answer]

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