One of two biased coins A and B is selected and flipped 3 times. Let A be the event that coin A is selected and B be the event that coin B is selected, with probabilities P(A) = 0.2 and p(B) = 0.8.
When coin A is flipped, the probability of heads is 0.9.
When coin B is flipped, the probability of heads is 0.5.
Let HHH be the event that the selected coin comes up heads 3 times. Complete the values X, Y, and Z in
Bayes' Theorem to determine the probability coin B was chosen if all 3 flips come up heads.
p(B|HHH)
Ex: 0.123
282) - 0.8
Ex: 0.123
-0.8 + Ex: 0.125
Y
Z
- 0.2

step-by-step explanation:

idk the answer i want to say 10 but then it would be equal so try 9 but then yeah so yeah we love edgen

we can't see all full 4 options, but if this is the full option, than the very left one is correct:

-2/5 * -3 3/4

remake the question, adding a picture of all the question, or write down the full left and right answers.

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.