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(a) A weighted coin hasprobability23 ofcoming up heads and probability13of coming up tails. The coin is tossed twice. Let x = number of heads. Set up the sample space for x and the associated probabilities.

(b) Find x and 蟽.

(c)If in (a) you know that there was at least one tail, what is the probability that both were tails?

Short Answer

Expert verified

(a)

P(x=0)=14P(x=1)=34P(x=2)=14

(b)

x=4/3,=2/3

(c)

PA(B)=1/5

Step by step solution

01

Given Information.

It has been given that a weighted coin has probability23 of coming up heads and probability13 of coming up tails

02

Definition of Probability. 

Probability means the chances of any event to occur is called it probability.

03

Find the probability.

a) The sample space S is given as shown below.

S={hh,ht,th,tt}

The variable x= "Number of heads" is given byx={0,1,2}

The associated probabilities are given as shown below.

P(x=0)=14P(x=1)=34P(x=2)=14

04

Find the variance. 

b) The mean valuexis given as shown below.

x=i=1nxip(xi)=019+149+249=43

The variance is given as shown below.

2=i=1n(xix)2p(xi)=(043)2(19)+(143)2(49)+(243)2(49)=(49)

Find the standard deviation.

=var(x)=(49)=23

05

Step 5:Find the probability. 

c) If we have a condition A= "that at least one of tosses is tail" and we need to find the probability of event B=" both are tails" then conditional probabilityPA(B) is given by the expression mentioned below.

PA(B)PA(B)=P(AB)P(A)=(13)(13)(13)(13)+2(23)(13)=(15)

Hence, the solutions are derived as:

(a)

P(x=0)=19P(x=1)=49P(x=2)=49

(b)

x=4/3,=2/3

(c)

PA(B)=1/5

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