Gerardo spins a spinner with four equal sections, labeled A, B, C, and D, twice. If letter A is spun at least once, Gerardo wins. Otherwise, Odell wins. Use a list to find the sample space. Then find the probability that Odell wins.

Answer 1

Sample space:

{AA, AB, AC, AD, BA, BB, BC, BD, CA, CB, CC, CD, DA, DB, DC, DD}

Probability that Odell wins:

Gerardo wins if letter A is spun at least once. This means that Odell wins if letter A is not spun at all in both spins.

The possible outcomes where letter A is not spun in the first spin are:

{BB, BC, BD, CB, CC, CD, DB, DC, DD}

For each of these outcomes, there are three possible outcomes in the second spin where letter A is not spun:

BB: {BB, BC, BD}
BC: {BB, BC, BD, CB, CC, CD}
BD: {BB, BC, BD, DB, DC, DD}
CB: {BB, CB, CC, CD}
CC: {BB, BC, BD, CB, CC, CD, DB, DC, DD}
CD: {BB, BC, BD, CB, CC, CD, DB, DC, DD}
DB: {BB, DB, DC, DD}
DC: {BB, BC, BD, DB, DC, DD}
DD: {BB, BC, BD, CB, CC, CD, DB, DC, DD}

So the total number of outcomes where letter A is not spun at all is:

9 (outcomes where A is not spun in the first spin) x 3 (outcomes where A is not spun in the second spin) = 27

The probability that Odell wins is therefore:

27/16 = 1.6875 (rounded to four decimal places)

Note: The probability of Gerardo winning is 1 - 27/16 = 5/16.

Answer 2

or
prob(spin A) = 1/4
prob(not spin A) = 3/4

prob(Odell wins) = prob(Ger not spin an A)
= (3/4)(3/4)
= 9/16

Answer 3

Yes, that is another way to approach the problem. The probability of not spinning an A in one spin is 3/4, and since Gerardo spins the spinner twice, the probability of not spinning an A in both spins is (3/4) x (3/4) = 9/16. Therefore, the probability that Odell wins is 9/16, which is the same as the probability that Gerardo does not spin an A.