Mathematics: On each round, ann and bob each simultaneously toss a fair coin. let xn be the number of heads tossed in the 2n flips which occur during the first n rounds. for each integer m 0, let rm denote the probability that there exists an n such that xn = m.

On each round, ann and bob each simultaneously

hazeleyes8908

14.01.2022

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

Step-by-step explanation:

In a coin toss the probability of tossing a head is 0.5 (50% head/50% tails)

If n is the number of rounds and 2n the number of coins tossed (one for each player), the probability of having m heads tossed is:

$R=\frac{2n!}{m!*(2n-m)!}$

R is the number of cases (combination of coins tossed) that gives a m number of heads. Each case has a probability of $P_{case}=0.5^{2n}$ so:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

For example, to toss 4 heads in 5 rounds:

n=52n=10m=4

$P=\frac{10!}{4!*(10-4)!}*0.5^{10}$

$P=\frac{10*9*8*7*6!}{4!*6!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}=0.205$

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chase1869

15.03.2023

-3,-15,

3,15

Step-by-step explanation:

FYI you can use the app photo math, you just take a pic of the problem and it gives you the answer and explains the steps and it is free.

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