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Natali [406]
3 years ago
11

A box contains identical balls of which 12 are red, 18 white and 8 blue. Three balls are drawn from the box one after the other

without replacement. Find the probability that;
(a) three are red
(b)the first is blue and the other two are red.​
Chemistry
1 answer:
svet-max [94.6K]3 years ago
8 0

Answer:

a) 12/323

b) 8/233

Explanation:

a) The probability of a red ball being drawn is 12/38, or in a simplified fraction, 6/19. To find the probability that 3 are red you would multiply the probability of the fraction for each, except subtracting one from the total each time as the drawn is done without replacement. This is done as follows: 6/19 × 6/18 × 6/17= 12/323

b) The probability of drawing a blue ball is 8/38, or 4/19. To find that the first one is blue and the rest are red, the equation is done as follows: 4/19 × 6/18 × 6/17 = 8/233

(hopefully I did this right)

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Where (E_{bond})_{i}  and (E_{bond})_{j} represents average bond energy in breaking "i" th bond and forming "j" th bond respectively.n_{i} and n_{j} are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

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