2% of the stray dogs in southfield are unlicensed. on a randomly-chosen day, the southfield city animal control officer picks up 7 stray dogs. what is the probability that at least on will be unlicensed?
0.98^7 is the probability that none of them are unlicenced. So, 1 minus this probability is the probability that one or more are unlicenced.
To calculate the probability that at least one stray dog will be unlicensed, we need to use the concept of complementary probability. The complementary event is the event that no stray dog is unlicensed.
Given that 2% of the stray dogs are unlicensed, the probability that a randomly chosen stray dog is licensed is 1 - 0.02 = 0.98.
Now, let's calculate the probability that all 7 stray dogs picked up by the animal control officer are licensed:
P(all licensed) = P(licensed) * P(licensed) * P(licensed) * P(licensed) * P(licensed) * P(licensed) * P(licensed)
= 0.98 * 0.98 * 0.98 * 0.98 * 0.98 * 0.98 * 0.98
≈ 0.86597
The probability that at least one stray dog will be unlicensed is equal to 1 - P(all licensed):
P(at least one unlicensed) = 1 - P(all licensed)
= 1 - 0.86597
≈ 0.13403
Therefore, the probability that at least one stray dog picked up will be unlicensed is approximately 0.13403 or 13.4%.