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Leviafan [203]
3 years ago
15

Plz help it is a probability question it is soo hard - I have tried and tried

Mathematics
1 answer:
zalisa [80]3 years ago
3 0
I think there's an easy way and a hard way to do this, and I think that the way
I'm about to describe is the easier way.

Probability = (number of successful outcomes)/(total number of possible outcomes)

<em>How many total pairs can be drawn from 8 total pens ?</em>
-- The first one drawn can be any one of 8 pens. For each of these . . .
-- The second one drawn can be any one of the remaining 7 .
-- Total number of ways of drawing a pair = (8 x 7) = 56 ways.
-- But there aren't 56 different different pairs. Whether you draw A and then B,
or B and then A, you wind up with the same pair.  There are 2 different ways to
draw each pair, so the 56 ways of drawing a pair only produces <u>28</u> different pairs.

<u>How many pairs are two of the same color ?</u>

<em>Possible number of blue pairs:</em>
The reasoning is exactly the same as calculating the TOTAL number of
pairs, as explained above.
With 5 blue pens, you can make <u>10</u> different pairs.
AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.

<em>Possible number of red pairs:</em>
The reasoning is exactly the same as calculating the TOTAL number of
pairs, as explained above.
With 3 red pens, you can make <u>3</u> different pairs.
AB, AC, and BC.

Total number of possible same-color pairs = 10 + 3 = 13

successes / total possible outcomes = 13/28 = <u>46.4</u>%  (rounded)
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Andre45 [30]

Answer:  A ∪ B = {a, b, c, d}

               (A ∪ B) ∪ C =  = {a, b, c, d, e}

               B ∪ C =  = {a, b, d, e}

               (B ∪ C) ∪ A  =  = {a, b, c, d, e}

<u>Step-by-step explanation:</u>

A = {a, b, c}       B = {a, b, d}         C = {b, d, e}

Union means "to join" so combine the sets to form a union.

A ∪ B = {A & B}

         = {a, b, c & a, b, d}

         = {a, b, c, d}          <em>because we do not need to list a & b twice</em>

(A ∪ B) ∪ C = {(A ∪ B) & C)

                  = {a, b, c, d & b, d, e}

                  = {a, b, c, d, e}     <em>because we do not need to list b & d twice</em>

<em />

B ∪ C = {A & B}

         = {a, b, d & b, d, e}

         = {a, b, d, e}           <em>because we do not need to list b & d twice</em>

<em />

(B ∪ C) ∪ A = {(B ∪ C) & A)

                  = {a, b, d, e & a, b, c}

                  = {a, b, c, d, e}     <em>because we do not need to list a & b twice</em>

<em />

<em />

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3 years ago
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