A baseball team is required to make four pitchers and six non-pitchers available

Question

A baseball team is required to make four pitchers and six non-pitchers available for an expansion draft. From this pool of 10 players the league selects two, with each player equally likely to be chosen first and each of the nine remaining players equally likely to be chosen second. Let A be the event that the first player chosen is a pitcher (and the second can be anything). Let B be the event that the second player chosen is a pitcher (and the first can have been anything). Express the event that, fit least one pitcher will be chosen in terms of A and B and evaluate its probability.

Explanation / Answer

There are four pitchers and six non pitchers.

A is the event that the first player chosen is a pitcher and B is the event that the second player chosen is a pitcher. If any of these events is satisfied, then atleast one pitcher will be chosen

Therefore the event can be expressed as A OR B or A U B.

The probability that the first player chosen is NOT a pitcher is 6/10. The probability that the second person chosen is also not a pitcher is 5/9. Thus the probability that neither player is a pitcher is 6/10*5/9 = 1/3.

=> Probability that atleast one pitcher is chosen = 1 - 1/3 = 2/3.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.