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.
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.
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
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.
To find the sample space for this problem, we need to list all possible outcomes for spinning the spinner twice. We can represent these outcomes using a list where each element represents a possible combination of two spins.
The sample space for spinning the spinner twice includes the following outcomes:
[AA, AB, AC, AD, BA, BB, BC, BD, CA, CB, CC, CD, DA, DB, DC, DD]
Now, we need to determine the probability that Odell wins. Odell wins if and only if the letter A is not spun in either of the two spins. Let's count how many outcomes satisfy this condition:
Out of the 16 possible outcomes in the sample space, only 15 outcomes do not have the letter A in both spins. Those outcomes are:
[BB, BC, BD, CB, CC, CD, DB, DC, DD]
The probability that Odell wins is given by the ratio of the number of favorable outcomes (9) to the total number of possible outcomes (16):
Probability(Odell wins) = Favorable outcomes / Total outcomes
= 9 / 16
= 0.5625
Therefore, the probability that Odell wins is 0.5625, or 56.25%.