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astra-53 [7]
3 years ago
6

Plzz help due in 2 mins

Mathematics
1 answer:
daser333 [38]3 years ago
5 0

Answer:

19.84

the decimal has to stay in the same place

Step-by-step explanation:

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G

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A store buys items at a wholesale price of $45 each. If the store marks up the price, and no discounts are given, which could be
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A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

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