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zavuch27 [327]
3 years ago
8

CONDITIONAL PROBABILITY URGENT?

Mathematics
1 answer:
kirill115 [55]3 years ago
6 0

Senior = 5

Senior but Girls = 4

so P(Senior | Girl) = 4/5

Hope it helps.

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8 0
3 years ago
The bucket containing orange tennis balls and yellow tennis balls from which 5 balls are selected at random, but assume that the
NeX [460]

(1)  

Probability that all are yellow = \frac{7\times6\times5\times4\times3}{16\times15\times14\times13\times12}=0.0048

Probability that all are orange = \frac{9\times8\times7\times6\times5}{16\times15\times14\times13\times12}=0.0288

Sum of the two probabilities = 0048+ 0.0288=0.0336

Thus, the probability that at least one is orange and at least one is yellow = 1-0.0336= 0.9664= 96.64 %


(2) In addition to the above, we eliminate the "1 and 4" cases.  

In these cases, any of the 5 can be taken out, so we multiply by 5.  


1 yellow + 4 orange = 5\times \frac{(7 \times 9 \times 8 \times 7 \times 6)}{(16 \times 15 \times 14 \times 13 \times12)}  = 0.2019

1 orange + 4 yellow = 5 \times \frac{(9 \times 7\times 6 \times 5 \times 4)}{(16 \times 15\times 14 \times 13 \times 12)} = 0.0721


Sum of two probabilities = 0.2019+0.0721=0.2740  

Now adding 0.0336 to this probability =0.0336+0.2740= 0.3076  


Subtracting that from 1.0000 = 1-0.3076=0.6924= 69.24 %

Thus, the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow= 69.24%


4 0
3 years ago
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