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4 years ago

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A bag of jelly beans contains 10 red,6 green,7 yellow, and 5 orange jelly beans.What is the probability of randomly choosing a r

ed jelly bean,replacing it,randomly choosing another red jelly bean,replacing it,and then randomly choosing an orange jelly bean?

1 answer:

lilavasa [31]4 years ago

7 0

Answer:

$\text{The Probability is = }\frac{125}{5488}\approx 0.02$

Step-by-step explanation:

No. of red jelly = 10

No. of green jelly = 6

No. of yellow jelly = 7

No. of orange jelly = 5

Total number of jelly = 10 + 6 + 7 + 5 = 28

$\text{So, Probability of choosing red jelly, P(R) = }\frac{10}{28}=\frac{5}{14}\\\\\text{So, Probability of choosing orange jelly, P(O) = }\frac{5}{28}\\\\\text{Now, Probability of choosing first red , then red and then orange jelly}\\\text{on replacing each jelly after choosing, }\\\\P(R,R,O) = \frac{5}{14}\times \frac{5}{14}\times \frac{5}{28}=\frac{125}{5488}\approx 0.02$

Hence, the required probability is 0.02 or 2 %

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$15(5)^2(-\frac{1}{2})^4$

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<em>From the above expression, the power of (5) is 2</em>

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$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

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$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

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Substitute values for a, b, n and r in

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Solve for $^6C_4$

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<em>Check the list of options for the expression on the left hand side</em>

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