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iren [92.7K]
2 years ago
5

PLease help my guys!

Mathematics
1 answer:
natima [27]2 years ago
5 0

Answer:

kinder

Step-by-step explanation:

1. 8

2. 2

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A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement
dsp73

Answer:

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

P(Same) = \frac{67}{210}

Step-by-step explanation:

Given (Omitted from the question)

Red = 7

White = 9

Blue = 5

Solving (a): P(First\ White\ and\ Second\ Blue)

This is calculated using:

P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)

P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}

P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}

P(First\ White\ and\ Second\ Blue) = \frac{45}{420}

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

Solving (b) P(Same)

This is calculated as:

P(Same) = P(First\ Blue\ and Second\ Blue)\or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)

P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})

P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})

P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}

P(Same) = \frac{20+42+72}{420}

P(Same) = \frac{134}{420}

P(Same) = \frac{67}{210}

6 0
3 years ago
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