In a jar of ten beads, seven are red and three are blue. A bead is drawn from the jar five times with replacement. What is the probability of seeing fewer than 2 blue beads?
Question 16 options:
52.82%
30.87%
36.02%
100%
None of these are correct.
To solve this problem, we need to consider the possible outcomes when drawing a bead from the jar five times. Since there are only two colors (red and blue), let's represent the colors as R and B, respectively.
To find the probability of seeing fewer than 2 blue beads, we need to calculate the probability of seeing 0 blue beads and 1 blue bead and then add them together.
1. Probability of seeing 0 blue beads:
In this case, all 5 beads must be red.
The probability of drawing a red bead on one try is 7/10.
Since we are drawing with replacement, the probability of drawing a red bead on each of the five tries is (7/10)^5.
2. Probability of seeing 1 blue bead:
In this case, there are three ways to get exactly one blue bead:
- The first draw is blue, and the remaining four draws are red
- The second draw is blue, and the remaining four draws are red
- The third draw is blue, and the remaining four draws are red
The probability of drawing a blue bead on one try is 3/10.
The probability of drawing a red bead on the remaining four tries is (7/10)^4.
Since there are three possible scenarios (with each having the same probability), we need to multiply by 3.
Therefore, the probability of seeing fewer than 2 blue beads is:
(7/10)^5 + 3 * (3/10) * (7/10)^4
= 0.16807 + 3 * 0.03087
= 0.16807 + 0.09261
= 0.26068
So the correct answer is not among the given options.