There is a .7 chance to select a toothbrush, .2 chance to select a yogurt, and .1 chance to select a monkey treat. There is a .8 chance the toothbrush will decay, .2 chance the yogurt will decay and .5 chance the monkey treat will decay. You select an item and it decays. What is the probability that is was the monkey treat?

To find the probability that the selected item was the monkey treat given that it decayed, we can use Bayes' theorem. Bayes' theorem states that:

P(A|B) = (P(B|A) * P(A)) / P(B),

where P(A|B) is the probability of event A occurring given that event B has occurred, P(B|A) is the probability of event B occurring given that event A has occurred, P(A) is the probability of event A occurring, and P(B) is the probability of event B occurring.

In this case, we want to find the probability that the selected item was the monkey treat (event A) given that it decayed (event B). Let's assign variables to the probabilities:

P(A) = 0.1 (chance to select a monkey treat)
P(B|A) = 0.5 (chance that the monkey treat will decay)
P(B) = ?

To find P(B), the probability that the selected item decays, we need to consider all the possible outcomes: toothbrush decaying, yogurt decaying, and monkey treat decaying.

P(B) = P(B|A) * P(A) + P(B|not A) * P(not A),

where P(not A) is the probability of not selecting a monkey treat.

P(not A) = 1 - P(A),
P(B|not A) = 0.8 * P(toothbrush) + 0.2 * P(yogurt),

where P(toothbrush) is the probability of selecting a toothbrush and P(yogurt) is the probability of selecting a yogurt.

P(toothbrush) = 0.7,
P(yogurt) = 0.2.

Let's calculate P(B|not A):

P(B|not A) = 0.8 * 0.7 + 0.2 * 0.2 = 0.56 + 0.04 = 0.6.

Therefore:

P(not A) = 1 - P(A) = 1 - 0.1 = 0.9.

Now we can calculate P(B):

P(B) = P(B|A) * P(A) + P(B|not A) * P(not A) = 0.5 * 0.1 + 0.6 * 0.9 = 0.05 + 0.54 = 0.59.

Finally, we can use Bayes' theorem to find the probability of selecting the monkey treat given that it decayed:

P(A|B) = (P(B|A) * P(A)) / P(B) = (0.5 * 0.1) / 0.59 = 0.05 / 0.59 ≈ 0.0847.

Therefore, the probability that the selected item was the monkey treat given that it decayed is approximately 0.0847, or about 8.47%.