Written lab instructions (example) to help you complete the lab:
Example: You are working with 25 M & Ms. Fill out the theoretical chart. Use the data
you took online for your empirical charts.
THEORETICAL probabilities use the theoretical chart. Suppose I have 5Y (yellow)\, 4G
(green)\, 2Bl (blue)\, 6B (brown)\, 5O (orange)\,and 3R (red); Total: 25 M & Ms}.
With replacement: You draw the first M & M\, put it back\, and draw the 2nd. For each
draw you have a total of 25 M & Ms to draw from.
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P(2 reds) = P(R and R) = (3/25)for the first draw multiplied by (3/25) for the
second draw = (3/25)(3/25). You can leave the answer in this form. You will
understand it better.
NOTE: R1 means red on the first draw. B2 means brown on the 2nd draw, etc.
P(R1B2 or B1R2) = P(R1 and B2) + P(B1 and R2)= (3/25)(6/25) + (6/25)(3/25)
P(R1 and G2) = (3/25)(4/25)
P(G2|R1) = P(R1 and G2)/P(R1) = (3/25)(4/25)/3/25) = 4/25
P(no yellows) means on either draw, you do not draw a yellow. There are 20 no
yellows. P(no yellows)= (20/25)(20/25)
P(doubles) = P(R and R) + P(G and G) + P(Bl and Bl) + P(B and B) + P(O and O)
+ P(Y and Y)
Doubles and no doubles are complements so P(doubles)+P(no doubles) = 1
Without replacement: You draw the first M & M, KEEP IT OUT,and draw the 2nd. Then
put BOTH M & Ms back to draw the next pair. For the first draw you have a total of 25
M & Ms and the second draw, 24 M & Ms.
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P(2 reds) = P(R and R) = (3/25) for the first draw multiplied by (2/24) for the
second draw = (3/25)(2/24). You can leave the answer in this form. You will
understand it better.
NOTE: R1 means red on the first draw. B2 means brown on the 2nd draw, etc.
P(R1B2 or B1R2) = P(R1 and B2) + P(B1 and R2) = (3/25)(6/24) + (6/25)(3/24)
P(R1 and G2) = (3/25)(4/24)
P(G2|R1) = P(R1 and G2)/P(R1) = (3/25)(4/24)/3/25) = 4/24
P(no yellows) means on either draw, you do not draw a yellow. There are 20 no
yellows. P(no yellows)= (20/25)(19/24)
P(doubles) = P(R and R) + P(G and G) + P(Bl and Bl) + P(B and B) + P(O and O)
+ P(Y and Y)
Doubles and no doubles are complements so P(doubles)+P(no doubles) = 1
For the EMPIRICAL probabilities for both with and without replacement, just count the
ordered pairs in the chart and divide by 24. The exception is P(G2|R1). Leave your
answers as fractions.
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P(2 reds) = P(R and R)= all ordered pairs that are (R,R) divided by 24 ordered
pairs. So, if there are 7 ordered pairs that are (R,R), the probability is 7/24.
P(G2|R1) = all ordered pairs (R,G) out of the order pairs that are (R,anything).
For example, suppose the (R,anything) ordered paires are (R,G), (R,G), (R,Y),
(R,B), (R,B), (R,Bl). There are 6 of these ordered pairs and 2 of them are (R,G) so
the probability is 2/6.
Make sure you answer the questions at the bottom of page 111 and on page 112 (page
112 in COMPLETE SENTENCES). When you convert to decimal, carry the answer to 4
decimal plac