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
Hope this helps mark brainliest please
Answer:
B. -1/8
Step-by-step explanation:
Using the terms given in the sequence:
4, -2, 1, -1/2, 1/4
we can find the pattern by dividing since it is a geometric seqence:
-2/4 = -1/2
1/-2 = -1/2
-1/2 ÷ 1 = -1/2
So, each term is multiplied by -1/2 to get the next term in the sequence:
It would be 16 more 6th graders that are not in either band or orchestra.
15 6th Graders in band
9 6th Graders in both classes
+18 6th Graders in orchestra
-----
42 6th Graders in either class
100 6th Graders in total
-42 6th Graders in either class
-------
58 6th Graders in neither class
58-42= 16 more 6th Graders that are not in either class
You're welcome :)
Answer:
Pls complete the question
Step-by-step explanation:
