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

Question

. 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) his first win (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? first win? including) his first win?

Explanation / Answer

The number of trials required to get one success could be represented as a geometric random variable.

If x follows geometric distribution with parameter p then ,the probability that the kth trial is the first success is

P( x = k ) = ( 1 - p)1-k p   , k = 1 , 2 , 3

Expected value = E( x) = 1/p

Variance = Var( x ) = ( 1 - p) / p2

We know person wins a game with probability 0.4

and x be the number of games that he must play until his first win.

so x follows geometric distribution with p = 40

a ) We have to find expected value.

Number of games he expect to play gam until the first success is,

E( x ) = 1/p = 1/0.4 = 2.5

b ) We have to find variance.

The variance of the number of games he plays until his first win is,

var( x ) = ( 1 - p)/p2 = ( 1 - 0.4)/(0.42)= 3.75

c)

We have to find P( x >= 4)

P( x >= 4) = 1 - P( x <= 3 )

P( x <= 3 ) = P( x = 1 ) + P( x = 2 ) + P( x = 3 )

Using formula of geometric distribution,

P( x = 1 ) = ( 1 - p )k-1p = ( 1 - 0.4 )1-1 (0.4) = 0.4

P( x = 2 ) = ( 1 - 0.4 )2-1 (0.4) = 0.24

P( x = 3 ) = ( 1 - 0.4 )3-1 (0.4) = 0.144

So, P( x <= 3 ) = 0.4 + 0.24 + 0.144 = 0.784

P( x >= 4) = 1 - 0.784 = 0.216

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