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YOU MAY BE WORKING WITH EITHER 25 OR 40 M & Ms. This example uses 25 M & Ms.
Use the data you took online as 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.
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 blue on the 2nd draw, etc.
P(R1B2 or B1R2) = P(R1 and B2) + P(B1 and R2)
= (3/25)(2/25) + (2/25)(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(R and R)
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.
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 blue on the 2nd draw, etc.
P(R1B2 or B1R2) = P(R1 and B2) + P(B1 and R2)
= (3/25)(2/24) + (2/25)(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(R and R)
Doubles and no doubles are complements so P(doubles)+P(no doubles) = 1
For the EMPIRICAL probabilities for both without replacement and with replacement, just count the ordered pairs in the chart and divide by 24. The exception is P(G2|R1). Leave answers as a fraction.
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.
When you convert to decimal, carry the answer to 4 decimal places. |
