1. Three people, A,B, and C, each flip their coins until one person has a differ

Question

1. Three people, A,B, and C, each flip their coins until one person has a

different result from the others. The person having the different coin

wins. For instance, if the three players flip HHH, TTT, HHT, C would

win.

(a) Simulate this experiment 10000 times and give the resulting estimate

of P(A wins).

(b) How mamy trials are necessary before the 95% confidence interval

has width .01?

(c) Interpret the statment “the 95% confidence interval has width .01”

in terms of the true value of P(A wins) and your interval.

(d) Argue for what you believe is the true probability of P(A wins).

2. A and B alternate drawing cards from a shuffled pack, replacing each card

when done. A goes first. Play continues until a heart is drawn. Simulate

this experiment to compute P(A draws first heart) to 2 decimal places

(±.005)

3. Two cards are drawn from a shuffled deck.

(a) Give the sample space for the experiment, , and calculate

||—

the number of elements in .

(b) How many elements of have both cards having the same rank? (i.e.

both aces or both kings etc.)

(c) What is the probability of drawing a pair? (both same rank)

4. An olympic archer hits the “bullseye” (the center of the target) half of the

time. Suppose the archer shoots 10 arrows. Compute:

(a) P(0 bullseyes)

(b) P(1 bullseyes)

(c) P(2 bullseyes)

(d) P(3 bullseyes)

Explanation / Answer

question 4

An olympic archer hits the “bullseye” (the center of the target) half of the

timso, probability of success p=0.5

here n=10

applying binomial distribution,

a)P(x=r)=C(n,r)*pr(1-p)(n-r)

P(x=0)=C(10,0)*(0.5)^0*(0.5)^10

=0.5^10

b)P(x=1)=C(10,1)*(0.5)^1*(0.5)^9

=10*0.5*0.5^9

=10*(0.5)^10

c)P(x=2)=C(10,2)*(0.5)^2*(0.5)^8

=45*(0.5)^10

d)P(x=3)=C(10,3)*(0.5)^3*(0.5)^7

=120*(0.5)^10

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