Question

Suppose Ari loses 31 ​% of all ping dash pong games . ​(a) What is the probability that Ari loses two ping dash pong games in a​ row? ​(b) What is the probability that Ari loses six ping dash pong games in a​ row? ​(c) When events are​ independent, their complements are independent as well. Use this result to determine the probability that Ari loses six ping dash pong games in a​ row, but does not lose seven in a row.

​(a) The probability that Ari loses two ping dash pong games in a row is . 0961 . ​(Round to four decimal places as​ needed.) ​(.31)^2

(b) The probability that Ari loses six ping dash pong games in a row is . 0009 . ​(Round to four decimal places as​ needed.) ​(.31)^6

(c) The probability that Ari loses six ping dash pong games in a​ row, but does not lose seven in a row is nothing . ​(Round to four decimal places as​ needed.)

Answer 1

solution:

Given Ari loses 31% of all ping dash pong games

P (Ari loses all ping dash pong games) = 0.31

a) Let L = event of lose of one ping dash pong games

P(L) = 0.31

W = event of one win ping dash pong games

P(W) = 1 - P(L) = 1 - 0.31 = 0.69

Remember L and W are independent

P(Ari loses 2 ping dash pong games ) = P( LL)

= P(L) * P(L)

= 0.31 * 0.31

= 0.0961

   Probability that Ari loses 2 ping dash pong games in a row = 0.0961

b) P(Ari loses 6 ping dash pong games ) = P(LLLLLL)

=   P(L) * P(L) * P(L) * P(L) * P(L) * P(L)

= 0.31 * 0.31 * 0.31 * 0.31 * 0.31 * 0.31

= 0.0009

   Probability that Ari loses 6 ping dash pong games in a row = 0.0009

c) P(Ari  loses 6 ping dash pong games but not 7 games ) = P(WLLLLLL) or P(LLLLLLW)

= P(L) * P(L) * P(L) * P(L) * P(L) * P(L) * P(W)

= 0.31 * 0.31 * 0.31 * 0.31 * 0.31 * 0.31 * 0.69

= 0.0006

    Probability that Ari loses 6 ping dash pong games ibut not 7 games in a row = 0.0006