# Math Help - Bernoulli, Binomial, and PMF questions

1. ## Bernoulli, Binomial, and PMF questions

I have some problems I have been working on. I have solved some and are stuck on some. I want to make sure that I am on the right track and not doing all of this completely wrong.

1) Suppose that visitors to a web page can click one of two buttons: “buy” or “cancel”. Visitors make decisions independently and for each visitor:
P[“cancel”] = 0.7
(a) If 6 people visit the web page, what is the probability that exactly two of them buy?
(b) If 6 people visit the web page, what is the probability that at least two of them buy?
(c) If 6 people visit the web page, what is the probability that only the last person buys?

a) I assumed that this was bernoulli's because there are only two options (buy and cancel). Thus since order does not matter, I came up with: (0.3)^2 * (1-0.3)^4 * (6!/(5!2!))
Which then solves out to be 0.3241

b) I am a little confused as to which equation I should be using for this. My professor gave us the Binomial Theorem, Geometric Prob Law, and the Multinomial Prob Law. I am a bit unsure as to which case this is. Here is what I did using the Binomial Prob Law:
n=5; k=2
10p^k(1-p)^(n-k)
(10)(0.3)^2 (0.7)^3 = 0.3087

I have a friend who did something like this but I am unsure if this is correct either:
1 - (0.7)^6 - 6*.3*(.7)^5

c) p^1 (1-p)^5
0.3(1-0.3)^5 * (6!/5!)
(6)(0.3)(0.7)^5 = 0.302526
Since we are only interested in finding the possibility for the last one, we would divide by 6 since the previous line is for w/o ordering.
0.302526/6 = 0.050421

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2) Suppose that an information source produces symbols from the set
S = {a, b, c, d, e},
with the following probabilities:
P[a] = 0.06, P[b] = 0.55, P[c] = 0.04, P[d] = 0.15, P[e] = 0.2 .
A data compression system encodes the letters as strings of binary digits as follows:
letter
string
a 1100
b 0
c 1101
d 111
e 10
Find the sample space and pmf of the following random variables.
(a) X = first digit of string
(b) Y = length of string
(c) Z = sum of digits in string

a) Sample Space {0,1}
pmf Px(0) = 1/5; Px(1) = 4/5

b) Sample Space {1,2,3,4}
pmf Px(1) = 1/5; Px(2) = 1/5; Px(3) = 1/5; Px(4) = 2/5

c)Sample Space {0,1,2,3}
pmf Px(0) = 1/5; Px(1) = 1/5; Px(2) = 1/5; Px(3) = 2/5

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3. Suppose that a discrete random variable X assumes values from the set SX = {0, 1, 2, …, 8}.
Suppose that the pmf of X has the form,
px(k) = a*k for k=0,1,2,...,8
and zero otherwise.
(a) Find the value of a .
(b) Find the probability that {X ≤ 4}.

a) I am very confused on this part but I came up with:
∑px(k) = 1
px(0)+px(1)+...+px(8) = 1
a = 1/36

b) I am clueless as to how to do this one... >_<

2. Hello,
Originally Posted by xicybluex
I have some problems I have been working on. I have solved some and are stuck on some. I want to make sure that I am on the right track and not doing all of this completely wrong.

1) Suppose that visitors to a web page can click one of two buttons: “buy” or “cancel”. Visitors make decisions independently and for each visitor:
P[“cancel”] = 0.7
(a) If 6 people visit the web page, what is the probability that exactly two of them buy?
(b) If 6 people visit the web page, what is the probability that at least two of them buy?
(c) If 6 people visit the web page, what is the probability that only the last person buys?

a) I assumed that this was binomial's because there are only two options (buy and cancel). Thus since order does not matter, I came up with: (0.3)^2 * (1-0.3)^4 * (6!/(4!2!))
Which then solves out to be 0.3241
Tiny typo that doesn't affect the result
(it's actually binomial, since there are successive Bernoulli events)

b) I am a little confused as to which equation I should be using for this. My professor gave us the Binomial Theorem, Geometric Prob Law, and the Multinomial Prob Law. I am a bit unsure as to which case this is. Here is what I did using the Binomial Prob Law:
n=5; k=2
10p^k(1-p)^(n-k)
(10)(0.3)^2 (0.7)^3 = 0.3087

I have a friend who did something like this but I am unsure if this is correct either:
1 - (0.7)^6 - 6*.3*(.7)^5
I don't really know how you got your formula, but your friend's is correct :
$P(X \geqslant 2)=1-P(X<2)=1-[P(X=0)+P(X=1)]$, and that's because X takes values into {0,1,2,3,4,5,6}. So if X<2, then it's equal to 0 or 1.

c) p^1 (1-p)^5
0.3(1-0.3)^5 * (6!/5!)
(6)(0.3)(0.7)^5 = 0.302526
Since we are only interested in finding the possibility for the last one, we would divide by 6 since the previous line is for w/o ordering.
0.302526/6 = 0.050421
Looks correct to me