Question

2. Jeff is trying to "get with" Slater. Being a calculating player, Jeff knows that his population proportion of landing a date is 69 percent. Jeff is going to make advances on Slater until success, so let us model the situation with a geometric distribution.2 (a) What is the probability that Jeff needs exactly 3 advances? (b) What is the probability that Jeff needs at least 3 advances? (c) Given that Jeff already made 4 advances, what is the probability that Jeff will need at least 3 more advances? (d) Show that all of the possibilities do indeed add up to 100 percent.

Answer 1

Answer:

$\bf a) - 0.0663\\b) - 0.31\\c) - 0.021\\d) - 1 or 100%$

Step-by-step explanation:

$Let\;p = 69\;\% =0.69$

$So, p(x)=q^{x-1}p$     $-(i)$

a) - The probability which Jeff requires at exactly three advances that is,

Then, put the value of $x=3$ in eq - i

$p(x=3)=(0.31)^{3-1}(0.69)$

$p(x=3)=0.0663$

b) - The probability which Jeff requires at least three advances that is,

$p(x\geq1)=1-p$

Put the value of $p$ in the eq

$p(x\geq 2)=1-0.69=0.31$

c) - Considering that he has only made four advances, the probability would be that he will require at least three more advances that is.

$p(\frac{x=7}{x\geq 4}) = \frac{p(x=7\;\cap\;x\geq 4)}{p(x\geq 4)}$

$p(\frac{x=7}{x\geq 4}) = \frac{p(x=7)}{1-p(x\leq 3)}$

Then, put the value in the eq - i

          $=\frac{(0.31)^{7-1}(0.69)}{1-\sum_{x=1}^{3}(0.31)^{x-1}(0.69)}$

          $=\frac{0.0006}{1-0.9702}$   $=0.0201$

d) - Display that all possibilities actually added up to 100% that is.

$p(x\geq1)=p(x=1)+p(x=2)+p(x=3)...........$

             $=q^{1-1}p+q^{2-1}p+q^{3-1}p$

by solving $x=1$ then, we get

            $=\frac{p}{1-q} =\frac{0.69}{1-031}$

So, we get 1 or 100%.

Answer 2

Answer:

A) 0.0066309 ; B) 0.0961 ; D) Proven

Step-by-step explanation:

Geometric Distribution : X ~ $q^(x-1) . p$

p = 0.69 ; q = 1- p = 1 - 0.69 = 0.31

A) P (x = 3) = $(0.31)^(3-1) (0.69) = 0.31^2 (0.69) = 0.0961(0.69)= 0.0066309$

B) P (x > 3) = 1 - [ P(x =1) + P(x = 2) ]

$1 - [(0.31)^0 (0.69) + (0.31)^1(0.69)] = 1 - (0.69 + 0.2139) = 1 - 0.9039 = 0.0961$

D) P (x > 1) = P (x =1) + P (x = 2) + P (x = 3) ...............

= p + pq + pq^2 + pq^3 ........ [Geometric series with a = p , r = q]

Sum of infinite geometric series = a / (1- r)

p / (1-q) = 0.69 / (1-0.31) = 0.69 / 0.69 = 1 {Hence Proven}