Answer:
probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3
Step-by-step explanation:
First of all;
Let B1 be the event that the card with two red sides is selected
Let B2 be the event that the
card with two black sides is selected
Let B3 be the event that the card with one red side and one black side is
selected
Let A be the event that the upper side of the selected card (when put down on the ground)
is red.
Now, from the question;
P(B3) = ⅓
P(A|B3) = ½
P(B1) = ⅓
P(A|B1) = 1
P(B2) = ⅓
P(A|B2)) = 0
(P(B3) = ⅓
P(A|B3) = ½
Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;
P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]
Thus;
P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]
P(B3|A) = (1/6)/(⅓ + 0 + 1/6)
P(B3|A) = (1/6)/(1/2)
P(B3|A) = 1/3
Answer:
Step-by-step explanation:
<em>Hey there!</em>
Well to add this we need to pu it in improper form.
7/5 + 23/4
Now we need to find the LCM.
5 - 5, 10, 15, 20, 25, 30
4 - 4, 8, 12, 16, 20, 24, 28
So the LCD is 20.
Now we need to change the 5 and 4 to 20.
5*4 = 20
7*4 = 28
<u>28/20</u>
4*5=20
23*5=115
<u>115/20</u>
Now we can add 28 and 115,
= 143/20
Simplified
7 3/20
<em>Hope this helps :)</em>
Answer:
10 cans
Step-by-step explanation:
4 blue gallons + 6 white gallons = 10 gallons total. each gallon fills one can. that's ten total cans
Answer:
b
Step-by-step explanation:
If you're doing probabilities, there is a 1:6 chance on both (im guessing dont trust me i did probabilities a long time ago)
