A box has 8 beads of the same size, but all are different colors. Tamy 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 Tamy would pick a red bead on the first draw, then a white bead, and finally a red bead again?

Question

03-07-2017
Mathematics

contestada

A box has 8 beads of the same size, but all are different colors. Tamy 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 Tamy would pick a red bead on the first draw, then a white bead, and finally a red bead again?

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texaschic101 texaschic101 · Answer 1 · 2017-07-03T14:22:19+02:00

texaschic101 texaschic101

03-07-2017

P(red bead) = 1/8
P(white bead) = 1/8
P(red bead) = 1/8

P(red,white,red) = 1/8 * 1/8 * 1/8 = 1/512 <==

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JeanaShupp JeanaShupp · Answer 2 · 2019-03-20T05:43:32+01:00

Answer: [tex]\frac{1}{512}[/tex]

Step-by-step explanation:

Given: The total number of beads of same size but different color in a box = 8

Since Tamy draws a bead randomly from the box, notes its color, and then puts the bead back in the box. Thus, all the events are independent.

The probability of drawing one bead (any colour) from the box [tex]=\frac{1}{8}[/tex]

Also, she repeats this 3 times and all the events are independent of each other.

Therefore, the probability that Tamy would pick a red bead on the first draw, then a white bead, and finally a red bead again =[tex]\frac{1}{8}\times\frac{1}{8}\times\frac{1}{8}=\frac{1}{512}[/tex]