A feed manufacturer produces two types of feed: for pigs and for fowls. 30% of the feed is for pigs and 70% is for fowls. 5% of the pig feed is expired and 10% of the fowl feed is expired. A bag of fe

Here's how to solve the problem using a tree diagram:

a) Sketch a tree diagram to show all possible outcomes and their respective probabilities.

Let P be the event that the feed is for pigs, and F be the event that the feed is for fowls.
Let E be the event that the feed has expired, and E' be the event that the feed has not expired.

Given probabilities:

• P(P) = 30% = 0.30
• P(F) = 70% = 0.70
• P(E | P) = 5% = 0.05
• P(E' | P) = 1 - 0.05 = 0.95
• P(E | F) = 10% = 0.10
• P(E' | F) = 1 - 0.10 = 0.90

The tree diagram is structured as follows:

Start
├─── Pig (P=0.30) ───┬─── Expired (P=0.05) ─── P(P and E) = 0.30 * 0.05 = 0.015
│                   └─── Not Expired (P=0.95) ── P(P and E') = 0.30 * 0.95 = 0.285
│
└─── Fowl (P=0.70) ───┬─── Expired (P=0.10) ─── P(F and E) = 0.70 * 0.10 = 0.070
                    └─── Not Expired (P=0.90) ── P(F and E') = 0.70 * 0.90 = 0.630

b) Find the probability that the bag of feed has expired.

To find the probability that a bag of feed has expired, we sum the probabilities of all paths leading to an expired bag.

The probability that the bag of feed has expired is $\boxed{0.085}$.