SOLUTION: Suppose that 2400 people are all playing a game for which the chance of winning is 49​%. Complete parts​ (a) and​ (b) below.
a. Assuming everyone plays exactl
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Question 1119350: Suppose that 2400 people are all playing a game for which the chance of winning is 49%. Complete parts (a) and (b) below.
a. Assuming everyone plays exactly five games, what is the probability of one person winning five games in a row?
P(five wins in a row)=________________
(Round to three decimal places as needed.)
On average, how many of the 2400 people could be expected to have a "hot streak" of five games?
_____________________
(Round to the nearest whole number as needed.)
b. Assuming everyone plays exactly ten games, what is the probability of winning ten games in a row?
P(ten wins in a row)=________________
(Round to five decimal places as needed.)
On average, how many of the 2400 people could be expected to have a "hot streak" of ten games?
__________________________________________
(Round to the nearest whole number as needed.)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
a. This is 0.49^5*=0.028
b. 68, if the probability from a is not rounded, 67 if it is.
c. This is 0.49^10, or 0.0008, which to three decimal places rounded is 0.001
d. 1.91 or 2 if not rounded 2.4 or 2 if rounded.
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