Mathematics: All that is left in a packet of candy are 7 reds, 4 greens, and 2 blues, what is the probability that a random drawing yields a green followed by a red assuming that the first candy drawn is put back into the packet?

Well I don t know where inequalities would come in but the third length would be solved as follows.a^2+b^2=c^210^2+18^2=424Answer;I hope this is correct, its what I would do if I had this equation....

All that is left in a packet of candy are 7 reds,

Ответ:

stodd9503

30.12.2021

Mathematics

Probability of drawing a green = 4/13
Probability of drawing a red = 2/13
probability of drawing a green and then a red is 4/13 * 2/13 = 6/169

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Ответ:

chrischris1

30.12.2021

Mathematics

Multiply 7(reds) by 4 (greens) to get 2/13

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Ответ:

abhibhambhani

06.12.2021

Mathematics

Well I don't know where inequalities would come in but the third length would be solved as follows.
a^2+b^2=c^2
10^2+18^2=424
$\sqrt{424}$

Answer; $2 \sqrt{106}$
I hope this is correct, its what I would do if I had this equation.

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All that is left in a packet of candy are 7 reds, 4 greens, and 2 blues, what is the probability that a random drawing yields a green followed by a red assuming that the first candy drawn is put back into the packet?

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