Suppose the Braves beat the Phillies in every game with probability 3/5.
a) What is the probability that in five particular games in a row, the Braves win exactly 3 games?
b) In a 'best of five' series, the two teams play games until one team first wins three games. What is the probability that the Braves win a 'best of five' series?
Can someone answer this with steps on how to do it? Please, it would be a great help!! :)
Answers are:
a) 216/625
b)2133/3125
a) What is the probability that in five particular games in a row, the Braves win exactly 3 games?
b) In a 'best of five' series, the two teams play games until one team first wins three games. What is the probability that the Braves win a 'best of five' series?
Can someone answer this with steps on how to do it? Please, it would be a great help!! :)
Answers are:
a) 216/625
b)2133/3125
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a) You need two things. First the probability that the Braves win EXACTLY 3 games, which is
(3 / 5) * (3 / 5) * (3 / 5) * (2 / 5) * (2 / 5) = 3^3 * 2^2 / 5^5
Second you need the number of outcomes in which the Braves win 3 times, that is, did they WWWLL or WLWLW (W = Win, L = Lose)? This is 5 choose 3 (if you know combinations, if you don't then see the note). 5 choose 3 is 5! / (3! * 2!) = 10. There are 10 ways for the Braves to win 3 games, and the probability of each one is 3^3 * 2^2 / 5^5, so
P(3 wins) = 10 * 3^3 * 2^2 / 5^5 = 2^3 * 3^3 / 5^4 = 216 / 625
b) Ok, this is trickier. Now WWWLL does not make sense because after WWW the games would have stoped. The possible outcomes are now
P(WWW) +
P(LWWW) + P(WLWW) + P(WWLW)
P(LLWWW) + P(WLLWW) + P(WWLLW) + P(LWLWW) + P(LWWLW) + P(WLWLW)
= 3^3 / 5^3 + 3 * (2 / 5 * 3^3 / 5^3) + 6 * (2^2 / 5^2 * 3^3 / 5^3)
= 5^2 * 3^3 / 5^5 + 5 * 2 * 3^4 / 5^5 + 2^3 * 3^4 / 5^5
= 2133/3125
NOTE: n chose k means n! / (k! * (n - k)!) and is the number of ways you can chose groups of size k from a collection of n elements (groups are non-ordered). For instance, the number of ways 4 people can pair up is 4! / 2! 2! = 6 (draw all combinations by hand and you'll see this is the answer). For how to derive the formula see http://en.wikipedia.org/wiki/Combination
(3 / 5) * (3 / 5) * (3 / 5) * (2 / 5) * (2 / 5) = 3^3 * 2^2 / 5^5
Second you need the number of outcomes in which the Braves win 3 times, that is, did they WWWLL or WLWLW (W = Win, L = Lose)? This is 5 choose 3 (if you know combinations, if you don't then see the note). 5 choose 3 is 5! / (3! * 2!) = 10. There are 10 ways for the Braves to win 3 games, and the probability of each one is 3^3 * 2^2 / 5^5, so
P(3 wins) = 10 * 3^3 * 2^2 / 5^5 = 2^3 * 3^3 / 5^4 = 216 / 625
b) Ok, this is trickier. Now WWWLL does not make sense because after WWW the games would have stoped. The possible outcomes are now
P(WWW) +
P(LWWW) + P(WLWW) + P(WWLW)
P(LLWWW) + P(WLLWW) + P(WWLLW) + P(LWLWW) + P(LWWLW) + P(WLWLW)
= 3^3 / 5^3 + 3 * (2 / 5 * 3^3 / 5^3) + 6 * (2^2 / 5^2 * 3^3 / 5^3)
= 5^2 * 3^3 / 5^5 + 5 * 2 * 3^4 / 5^5 + 2^3 * 3^4 / 5^5
= 2133/3125
NOTE: n chose k means n! / (k! * (n - k)!) and is the number of ways you can chose groups of size k from a collection of n elements (groups are non-ordered). For instance, the number of ways 4 people can pair up is 4! / 2! 2! = 6 (draw all combinations by hand and you'll see this is the answer). For how to derive the formula see http://en.wikipedia.org/wiki/Combination