Morgan rolls a number cube, twice. If the number 1 shows up at least once, Morgan wins. Otherwise, Jaclyn wins. How much greater is the probability that Morgan will win compared to Jaclyn winning?
A)1/3 B)7/18 C)4/9 D)2/3
1 2 3 4 5 6 first die
1 1 1 1 1 1 second die
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
total of 36 possibles
the number one shows up 11 times
36 -11 = 25 times no ones
11/36 -25/36 = - 14/36 = -7/18
To find the probability that Morgan will win, we need to calculate the complementary probability of Jaclyn winning. The complementary probability of an event is equal to 1 minus the probability of that event occurring.
Let's calculate the probability of Jaclyn winning first:
There are 6 possible outcomes for each roll of the number cube (numbers 1 to 6). Since we are rolling the number cube twice, there are 6 * 6 = 36 equally likely outcomes.
Now, let's find the probability that no 1 shows up on both rolls:
On the first roll, the probability of not getting a 1 is 5/6 (since there are 5 possible outcomes that are not 1 out of 6). Similarly, on the second roll, the probability of not getting a 1 is also 5/6.
To find the probability of both events happening (no 1 on the first roll and no 1 on the second roll), we multiply the probabilities together:
(5/6) * (5/6) = 25/36
Therefore, the probability of Jaclyn winning is 25/36.
Now, let's find the probability of Morgan winning:
The probability of Morgan winning is equal to 1 minus the probability of Jaclyn winning:
1 - 25/36 = 11/36
To find out how much greater the probability that Morgan will win is compared to Jaclyn winning, we subtract the probability of Jaclyn winning from the probability of Morgan winning:
(11/36) - (25/36) = -14/36 = -7/18
Since probability cannot be negative, we take the absolute value of the result, which is 7/18.
Therefore, the answer is B) 7/18.