If the probability that the Islanders will beat the Rangers in a game is 85%, what is the probability that the Islanders will win exactly three out of five games in a series against the Rangers? Round your answer to the nearest thousandth.
3(.85) + 2(.15) = ?
The probability of an event, p, occurring exactly r times:
n Cr * p^r * q^(n-r)
Binomial Probability-1
n = number of trials
r = number of specific events you wish to obtain
p = probability that the event will occur
q = probability that the event will not occur
(q = 1 – p, the complement of the event)
p = .85
q = .15
n = 5
r = 3
C(n,r) = n!/ [ r!(n-r)! ] = 5!/[ 3!( 2!) ] = 5 *4 /2 = 10
so
10 * .85^3 * .15^2
= 0.138
2 ways to interpret the question:
1. There are actually 5 games
2. What is the prob that the Islanders win 3 games to win the series.
1. In the 5 games for the Islanders to win the series, the last game must
be a win, W, the other 4 games must be permutations of WWLL
---- there are 4!/(2!2!) or 6 of these
Prob(Islanders win in exactly 5 games) = 6(.85)^3 (.15)^2 = .0829
2. Possible ways to win series Islanders can win:
a) win in 3 straight --- .85^3 = .614125
b)) win in 4 games, there are 3 such cases, only 1 loss, one is WWLW
prob(to win in 4 games) = 3(.85)^3(.15) = .27636
c) win in 5 games, see above 1)
prob(t win in 5 games) = .0825
prob(islander win the series ) = .614125 + .27636 + .0825 = .973
(notice that this does not add up to 1, since there is a chance
the Islanders could lose the series)
I am going to scream.
To find the probability that the Islanders will win exactly three out of five games in a series against the Rangers, we can use the binomial probability formula:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
P(X = k) represents the probability of getting exactly k successes (in this case, winning games),
n represents the total number of trials (in this case, the total number of games in the series),
k represents the number of successes we are looking for (in this case, winning exactly three games),
p represents the probability of success on a single trial (in this case, the probability of the Islanders winning a game), and
(1-p) represents the probability of failure on a single trial (in this case, the probability of the Islanders losing a game).
Given that the probability of the Islanders beating the Rangers in a game is 85%, we can express this as a decimal as p = 0.85. Therefore, the probability of losing a game would be (1-p) = (1-0.85) = 0.15.
Now let's calculate the probability using the formula:
P(X = 3) = (5C3) * 0.85^3 * 0.15^(5-3)
To calculate (5C3), we can use the combination formula:
(nCk) = n! / (k! * (n-k)!)
(5C3) = 5! / (3! * (5-3)!)
= 5! / (3! * 2!)
= (5 * 4 * 3!) / (3! * 2)
= (5 * 4) / 2
= 10
Substituting these values into the binomial probability formula:
P(X = 3) = 10 * 0.85^3 * 0.15^2
Calculating this, we get:
P(X = 3) = 10 * 0.614125 * 0.0225
= 0.13853
Therefore, the probability that the Islanders will win exactly three out of five games in a series against the Rangers, rounded to the nearest thousandth, is approximately 0.139.