Question

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. 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

Answer 1

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 = $\frac{1}{5} \times \frac{1}{5} \times \frac{1}{5}$

                                              = $\frac{1}{125}$

Answer 2

Answer:

answer= d

1/125

Step-by-step explanation:

i got it right o the test :)