You roll a number cube twice. Find P (even, then not 2). Write the probability as a fraction in simplest form. plaz help
1/6 is p(2) ... p(not 2) is 5/6
whoa, dudes chill it...
school end tomorrow at one, i have no idea what is going on with this question, ive been staring at it for an hour, its so worded so damn bad, you cant tell whats going on. This question is like brawl meta knight, impossible to beat.
There are six possible outcomes when rolling a number cube: 1, 2, 3, 4, 5, or 6. Exactly half of these outcomes are even: 2, 4, 6. So, the probability of rolling an even number on the first roll is 1/2.
Now, we want to find the probability of rolling a number that is even on the first roll AND not 2 on the second roll. There are three even numbers, two of which are not 2. So, the probability of rolling a number that is even AND not 2 is (2/3) x (1/6) = 1/9.
Therefore, P(even, then not 2) = 1/2 x 1/9 = 1/18.
Answer: 1/18.
yeah Idc Ima just call my teachers to get my grades a bit higher.
hewlp
You roll a number cube twice. Find
P
(
even, then not
2
)
. Write the probability as a fraction in simplest form.
Scott is right of course, I am in "sloppy thinking" mode again.
Prob(even) = 3/6 = 1/2
prob(not 2) = 1/6
so .....