Question

3. In a certain game of chance, you have a 3/4 chance of winning each time you play with the outcomes each time you play independent of each other). Suppose you play the game until you win for the first time. What is the probability you will win the game on the first second, or third time playing it? 4. A box of 40 fuses contains 10 fuses which are defective. If you randomly choose a collection of 5 fuses from the box, what is the probability that 0 or 1 of the 5 fuses in your collection will be defective?

Answer 1

3.

X: Number of trails for the first win

Probability of winning : p =3/4

X follows a geometric distribution with p=3/4=0.75

Probability mass function : Probability that you will the game on 'x' th time playing it given by

P(X=x) = (1-p)^x-1p = (1-0.75)^x-1 x 0.75 = 0.25^x-1 x 0.75 for x=1,2,3,....

Probability that you will the game on the first time playing it = P(X=1)   = 0.25^1-1 x 0.75 =0.75

Probability that you will the game on the second time playing it = P(X=2)   = 0.25^2-1 x 0.75 = 0.25 x 0.75= 0.1875

Probability that you will the game on the third time playing it = P(X=3)   = 0.25^3-1 x 0.75 = 0.25² x 0.75= 0.046875

Probability that you will the game on the first, second, or third time playing it

= P(X=1)+P(X=2)+P(X=3) = 0.75+0.1875+0.046875=0.984375

Probability that you will the game on the first, second, or third time playing it = 0.984375

4.

Number of fuses in the box = 40

Number of defective fuses = 10

Number of randomly chosen collection of  fuses = 5

Number of ways of choosing collection of 5 fuses from 40 fuses = $\small \binom{40}{5}$ =658008

Probability that 0 of the 5 fuses in the collection will be defective i.e that all 5 fuses in the collection will be non defective

= Number of ways of selecting 5 fuses from the 30 non-defective fuses / Number of ways of choosing collection of 5 fuses from 40 fuses

Number of ways of selecting 5 fuses from the 30 non-defective fuses = $\small \binom{30}{5}$ =142506

Probability that 0 of the 5 fuses in the collection will be defective i.e that all 5 fuses in the collection will be non defective

= Number of ways of selecting 5 fuses from the 30 non-defective fuses / Number of ways of choosing collection of 5 fuses from 40 fuses

= 142506/658008=0.216571835

Probability that 1 of the 5 fuses in the collection will be defective i.e 1 fuse is defective and 4 non-defective fuses in the collection

=Number of ways of 1 defective fuse from 10 defective fuses x Number of ways of selecting 4 fuses from the 30 non-defective fuses / Number of ways of choosing collection of 5 fuses from 40 fuses

Number of ways of 1 defective fuse from 10 defective fuses = 10

Number of ways of selecting 4 fuses from the 30 non-defective fuses = 27405

Probability that 1 of the 5 fuses in the collection will be defective i.e 1 fuse is defective and 4 non-defective fuses in the collection

=Number of ways of 1 defective fuse from 10 defective fuses x Number of ways of selecting 4 fuses from the 30 non-defective fuses / Number of ways of choosing collection of 5 fuses from 40 fuses

=(10 x 27405)/658008 = 0.416484298

Probability that 0 or 1 of the 5 fuses in your collection will be defective

= Probability that 0 of the 5 fuses in the collection will be defective + Probability that 1 of the 5 fuses in the collection will be defective

= 0.216571835+0.416484298=0.633056133

Probability that 0 or 1 of the 5 fuses in your collection will be defective = 0.633056133