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alukav5142 [94]
2 years ago
15

PLEASE HELP! Will give brainliest! A bag contains 5 pink marbles and 6 green marbles. Two marbles will be selected one at a time

. The first marble a
iS
pink. This marble is placed on a table. The second marble is green.
Select words from the drop-down menus to correctly complete the sentences.
The selection of the two marbles are dependent or independent
because the selection of the pink marble
changes or doesn’t change
the probability that
green
marble iS drawn next. If the pink marble had been put back into
the bag, the selection of the two marbles would have been dependent or independent

Mathematics
1 answer:
Karolina [17]2 years ago
5 0

Answer:

The selection of the two marbles are <u>dependent</u> because the selection of the pink marble <u>changes</u> the probability that a green marble is drawn next.

If the pink marble had been put back into the bag, the selection of the two marbles would have been <u>independent</u>

======================================================

Explanation:

The key phrase to focus on here is "This marble is placed on a table". It implies that we do not put it back into the bag, and also implies we don't put in a replacement either. Hence we consider this scenario as "no replacement".

In no replacement situations, we always have dependent events. The second selection depends on the first one.

The probability of picking green is 6/11 because there are 6 green out of 5+6 = 11 total. But if we picked a pink one first and didn't put it back, then the 11 drops to 10 and we get 6/10 = 3/5 as the new probability.

In other words,

  • before the pink marble is selected, the chances of green are 6/11
  • after the pink marble is selected, and not put back, the chances of green are now 6/10 = 3/5

Because this probability changes, it's sufficient evidence to say the event of picking green on the second selection is dependent on the first selection of selecting pink.

If the marble had been put back, then we'd revert back to the original state of the bag. That would mean the chances of green would stay at 6/11. Therefore, situations with replacement lead to independent events.

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Let's expand both options, your friend's and yours, as shown below

\begin{gathered} y=(3x+1)(3x-5)=9x^2-12x-5=9(x^2-\frac{4}{3}x-\frac{5}{9}) \\ \text{and} \\ y=(x+\frac{1}{3})(x-\frac{5}{3})=x^2-\frac{4}{3}-\frac{5}{9} \end{gathered}

Then, both equations are the same besides a constant that will not affect the zeros of the functions, as shown below

\begin{gathered} y=(3x+1)(3x-5) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-1+1)(-1-5)=0 \\ \text{and} \\ x=\frac{5}{3} \\ \Rightarrow y=(5+1)(5-5)=0 \\  \end{gathered}

And

\begin{gathered} y=(x-\frac{5}{3})(x+\frac{1}{3}) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-\frac{1}{3}-\frac{5}{3})\cdot0=0 \\ x=\frac{5}{3} \\ \Rightarrow y=0\cdot(\frac{6}{3})=0 \end{gathered}

Both your friend and you are correct. The functions are the same with exception of a constant that multiplies the whole function (a scale factor); despite that, the zeros are the same for both functions

(3x-5)(3x+1)=9(x+\frac{1}{3})(x-\frac{5}{3})

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Answer:

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However, once you're looking at P(A and B) then you're fixing the first position and saying if the first student is a girl what's the probability of the second student being a girl.

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