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Binomial distribution

Can someone help me with this plss

A door to door canvasser tries to persuade people to have a certain type of double glazing installed. The probability that his canvassing at a house is successful is 0.05.

Calculate the least number of houses that he must canvass in order that the probability of his getting at least one success exceeds 0.99.
Reply 1
Avatar for Mawar3
Mawar3
1
Go back to your very first knowledge of the binomial distribution:
P(X=1)=nC1*(P(X))^1)*(P(X')^(n-1))
I.e. probability that one person agrees equals= number of different arrangements of one success and lots of not successes* [probability of this one success *probability of the lots of not successes]

[-]= probability of each arrangement

Does this help?
Reply 2
Avatar for BabyMaths
BabyMaths
Original post by fifi97
Can someone help me with this plss

A door to door canvasser tries to persuade people to have a certain type of double glazing installed. The probability that his canvassing at a house is successful is 0.05.

Calculate the least number of houses that he must canvass in order that the probability of his getting at least one success exceeds 0.99.


"at least one" means "not none".

If X∼B(n,0.05)X \sim B(n,0.05) you need to consider P(X≥1)=1−P(X=0)>0.99P(X \ge 1)= 1-P(X=0)>0.99.
(edited 9 years ago)
Original post by fifi97
Can someone help me with this plss

A door to door canvasser tries to persuade people to have a certain type of double glazing installed. The probability that his canvassing at a house is successful is 0.05.

Calculate the least number of houses that he must canvass in order that the probability of his getting at least one success exceeds 0.99.


Hmm I seen this exact problem in a top Singapore JC's lecture notes.

Anyways, let X denote the number of successes the canvasser encounters in knocking of the doors of n number of houses.

Then X~B(n, 0.05)

Then P(X ≥ 1) > 0.99

1-P(X=0) > 0.99 ---------------(1)

What does P(X=0) mean? The canvasser failed in securing any success in all n houses visited. As such, convince yourself P(X=0) = (1-0.05) ^n = 0.95^ n

Substitute this into (1), and you should be able to continue thereafter.

Peace.
(edited 9 years ago)
Original post by BabyMaths
"at least one" means "not none".

If X∼B(n,0.95)X \sim B(n,0.95) you need to consider P(X≥1)=1−P(X=0)>0.99P(X \ge 1)= 1-P(X=0)>0.99.


It should be X∼B(n,0.05)X \sim B(n,0.05) . Peace.
Reply 5
Avatar for BabyMaths
BabyMaths
Original post by WhiteGroupMaths
It should be X∼B(n,0.05)X \sim B(n,0.05) . Peace.


Ah, thanks, I should have gone to bed. I was thinking about his far more talented colleague. I'll edit my post.
I like your answer lol. Peace.
Reply 7
Avatar for fifi97
fifi97
OP
Thankyou all very much!! lol, got the answer before I even checked YAY :tongue:

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