Abox has 5 beads of the same size, but all are different colors. tina draws a bead randomly from the

8 answers
Question:

Abox has 5 beads of the same size, but all are different colors. tina draws a bead randomly from the box, notes its color, and then puts the bead back in the box. she repeats this 3 times. what is the probability that tina would pick a red bead on the first draw, then a green bead, and finally a red bead again? 1 over 625 1 over 180 1 over 150 1 over 125

Answers

Probability of first red = 1/5
and Probabiliiy  of the green =1/5 the red again is 1/5. 

Required probability is (1/5)^3  = 1/125

its D

Give the other gut

Step-by-step explanation:

5 beads each a different color

 so every time she picks a bead she has a 1/5 probability of picking a certain color

she does it 3 times so you get:

1/5 x 1/5 x 1/5 = 1/125 probability of picking those colors

[tex]\frac{1}{25}[/tex]

Step-by-step explanation:

To solve this problem we have the standard definition of a probability, that is:

[tex]P=\frac{n\° cases}{n\° total events}[/tex]

So, we have to identify how many cases are about the probability that it's being asked, and then we divide it by the total number of events possible.

In this case, each bead has the same probability of being picked, that is, [tex]\frac{1}{5}[/tex], because we have 5 beads in total, and only one of each color. So, all beads have [tex]\frac{1}{5}[/tex] of probability to be selected.

In addition, Tina wants to reproduce the experiment three times, where she puts back each bead after being picked, this means that each events is independent, that is, each bead doesn't influence the probability of others. When events are independent, this is shown as a multiplication, so basically we have to multiply three times the same fraction.

[tex]P = \frac{1}{5} \frac{1}{5} \frac{1}{5} = \frac{1}{125}[/tex]

Therefore, the probability of picking first a red bead, second a green one and third a red one agains, is [tex]\frac{1}{125}[/tex], which is a pretty low chance, it's hard to happen.

the answer is 1 over 125

Step-by-step explanation:

Probability of picking red on first draw = 1/5
but then the bead is put back
probability of picking green = 1/5
but then the bead is put back
probability of picking red again = 1/5

1/5 * 1/5 * 1/5 = 1/125 <===

The answer is 1 over 125

Required Probability = 1 over 125 (1/125)

Step-by-step explanation:

We are given that a box has 5 beads of the same size, but all are different colors. Tina draws a bead randomly from the box, notes its color, and then puts the bead back in the box. She repeats this 3 times.

We have to find the probability for the case when Tina pick a red bead on the first draw, then a green bead and finally a red bead again.

We have to first note here is that this a case of with replacement situation.

Probability formula is given by = \frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}

Now, Probability of picking a red bead on first draw

                           = \frac{\text{Number of red beads in a box}}{\text{Total number of beads in box}} = \frac{1}{5}

Similarly, Probability of picking a green bead on second draw

                           = \frac{\text{Number of green beads in a box}}{\text{Total number of beads in box}} = \frac{1}{5}

Probability of picking a red bead on again on third draw

                           = \frac{\text{Number of red beads in a box}}{\text{Total number of beads in box}} = \frac{1}{5}

Here, after picking each bead the number of beads available for another draw will be 5 only because Tina draws a bead randomly from the box, notes its color, and then puts the bead back in the box. So, number of total beads remained 5 only.

Hence, required probability = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5}

                                              = \frac{1}{125}

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