1. Find the expectation and variance of Y, the number of failures, in n binomial

Question

1. Find the expectation and variance of Y, the number of failures, in n binomial trials with p as the probability of success.

2. A bird cage has three healthy plump canaries and two scrawny ones. after a struggle Felix, the cat, nabs the three healthy birds and prepares to goblle them. just at that moment his master appears and says, "Felix, repeat that feat twice in three tries and you'll have the birds for dinner." What is the probability that Felix has the birds for dinner?

3. In a game of chance John tosses a balanced coin a fixed number of times. He receives a prize if he gets exactly five heads. Before tossing the coin, he is allowed to choose the number of trials. what number should he choose in order to maximize his probability of winning?

Explanation / Answer

1. let X denotes the number of success in n binomial trials with p as the probability of success.

then X~Bin(n,p)

Y denotes the number of failures

so Y=n-X

now E[X]=np and V[X]=np(1-p)

so expectation of Y is E[Y]=E[n-X]=n-E[X]=n-np=n(1-p)

variance of Y is V[Y]=V[n-X]=V[X]=np(1-p)

2. there are all total 5 birds out of which 3 are healthy and 2 are scrawny.

let nabbing the 3 healthy ones be denoted as success.

p=probability of success=choosing 3 healthy birds out of 3 healthy and 2 scrawny birds=3C3/5C3=1/10

if felix repeat that feat twice in three tries he will have birds for dinner.

let X denotes the number of times felix repeat that feat in 3 tries with probability of success=1/10

so X~Bn(3,1/10)

the probability that Felix has the birds for dinner=P[X=2]=3C2(1/10)2(1-1/10)3-2=3*9/1000=27/1000=0.027 [answer]

3. let n be the number of trials. the coin is balanced hence probability of head is 0.5

let X denotes the number of times head appears.

so X~Bin(n,0.5)

he receives a prize if he gets exactly 5 heads

we need to find n in order to maximize his probability of winning

so we need to find that n for which x=5 is the mode.

now for binomial disribution

if (n+1)p is a non integer then [(n+1)p] is the mode where [(n+1)p] is the greatest integer less than or equal to (n+1)p

if (n+1)p is an integer then (n+1)p and (n+1)p-1 are the two modes

now we have mode=5

so we have either (n+1)p=5 or (n+1)p-1=5

here p=probability of success=0.5

so (n+1)*0.5=5 or, n+1=10 or n=9

(n+1)0.5-1=5 or, n+1=12 or n=11

and if (n+1)p is non integer

then [(n+1)p]=5

or (n+1)p=5+k where 0<k<1

or, (n+1)*0.5=5+k

or, n+1=10+2k

or, n=9+2k

so 9+2k is an integer . so k has to be either 0 or 0.5 but assumption is 0<k<1

so k=0.5

hence n=9+2*0.5=10

so we have 3 choices of n. 9,10 and 11

which means for these 3 distributions Bin(9,0.5),Bin(10,0.5),Bin(11,0.5) x=5 is mode

now P[X=5 where X~Bin(9,0.5)]=9C5(0.5)9=0.2460938

P[X=5 where X~Bin(10,0.5)]=10C5(0.5)10=0.2460938

P[X=5 where X~Bin(11,0.5)]=11C5(0.5)11= 0.2255859

so P[X=5] is maximized when n=10 or 9

so he must toss the coin either 9 times or 10 times

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