In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.
To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9
To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum). The probability is 11/36
Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.
P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9
Answer:
the answer is C
Step-by-step explanation:
Let's solve for "r" by bringing them all to one side.
-3r=6-4r
+4r +4r
r=6
So, r is equal to 6.
Answer:JKL
Step-by-step explanation:A 270 degree rotation represents three 90 degree rotations
Rotating ABC by 90 degrees three times places it in the position of JKL
So, the answer is B. JKL
42/45 with in simplest form would equal to 1 14/42