4. Suppose that a person wins a game of chance with probability 0.40, and loses

Question

4. Suppose that a person wins a game of chance with probability 0.40, and loses otherwise. He plays the game until he wins for the first time, and then stops. Assume that the games are independent of each other. Let X denote the number of games that he must play until (and including) (a) How many games does he expect to play until (and including) his (b) What is the variance of the number of games he plays until (and (c) What is the probability that he plays 4 or more games altogether? his first win. first win? including) his first win?

Explanation / Answer

a)

here this is geometric distribution with parametert p=0.4

hence mean value =1/p =1/0.4 =2.5

b)variance =(!-p)/p2 =(1-0.4)/(0.4)2 =3.75

c)P(he plays 4 or more games) =P(no success in first 3 games) =(0.6)3 =0.216

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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