On the ninth day of Statistics, my teacher gave to me --- nine ladies dancing. These nine ladies dancing are the backup dancers for In Sync. It is well known that if dancers don’t warm up adequately before dancing, the dancers will be Out of Sync and may sustain injuries. Of the nine ladies, 1 warmed up and had injuries, 1 failed to warm up and did not have injuries, 5 warmed up and had no injuries and 2 failed to warm up and had injuries,. What is the probability that a randomly selected dancer would have injuries or would have failed to warm up?
9 dancers total
5 warmed and had no injuries (W & ~I)
You want ~W or I = ~(W & ~I)
So, your probability is 1 - 5/9 = 4/9
Thanks
To find the probability that a randomly selected dancer would have injuries or would have failed to warm up, we can use the principle of inclusion-exclusion.
Let's assign the following events:
I = The dancer has injuries
W = The dancer failed to warm up
We are asked to find P(I ∪ W), the probability of the union of the events "dancer has injuries" and "dancer failed to warm up."
To calculate this probability, we need to find the individual probabilities of each event and subtract the probability of their intersection.
P(I ∪ W) = P(I) + P(W) - P(I ∩ W)
From the given information, we know the following:
- The probability of injuries, P(I), is 1 out of 9 dancers (1 warmed up and had injuries).
- The probability of not warming up, P(W), is 2 out of 9 dancers (2 failed to warm up and had injuries).
- The probability of both injuries and not warming up, P(I ∩ W), is 0 out of 9 dancers (no one falls into this category).
Substituting the values into the formula:
P(I ∪ W) = P(I) + P(W) - P(I ∩ W)
= 1/9 + 2/9 - 0/9
= 3/9
= 1/3
Therefore, the probability that a randomly selected dancer would have injuries or would have failed to warm up is 1/3.