Question

In: Statistics and Probability

Suppose you have a pair of tetrahedra. One is red on one face, yellow on two...

Suppose you have a pair of tetrahedra. One is red on one face, yellow on two faces, and green on one face. The other is white and has faces marked 1, 2, 3 ,4

a. Complete the table

	1	2	3	4
Red
Yellow
Yellow
Green

b. If both tetrahedra are tossed, what is the probability of a red (facing down) and a 3 (facing down)? Of a yellow (facing down) and a number >1 on the other (facing down?) Of a green (facing down) and a number >4 (facing down) on the other? Of a yellow (facing down) on the colored one and a sum of >2 of faces showing on the other?

Expert Solution

	1	2	3	4	Total
Red	0.0625	0.0625	0.0625	0.0625	0.25
Yellow	0.0625	0.0625	0.0625	0.0625	0.25
Yellow	0.0625	0.0625	0.0625	0.0625	0.25
Green	0.0625	0.0625	0.0625	0.0625	0.25
Total	0.25	0.25	0.25	0.25	1

If both tetrahedra are tossed, what is the probability of a red (facing down) and a 3 (facing down)?

From the above table the P( a red (facing down) and a 3 (facing down) ) = 1/16 = 0.0625

Of a yellow (facing down) and a number >1 on the other (facing down?)

P( a yellow (facing down) and a number >1 on the other (facing down?) )

Let Y = Yellow

Therefore required probability = P( Y, 2) + P( Y, 2 ) + P( Y, 3) + P( Y, 3) + P( Y, 4) + P( Y, 4) = 6*(0.0625) = 0.375

Of a green (facing down) and a number >4 (facing down) on the other?

P( a green (facing down) and a number >4 (facing down) on the other) = 0

Because this is an empty event. ( there is no greater than 4 ).

Of a yellow (facing down) on the colored one and a sum of >2 of faces showing on the other?

P( a yellow (facing down) on the colored one and a sum of >2 of faces showing on the other ) =

Note that P( Yellow) = 8/16 = 0.5

And there are only one posibility of Yellow and sum = 2

So required probability = (8/16) - -(1/16) = 7/16 = 0.4375

orchestra answered 2 years ago

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