A die has 6 sides
The probability of getting a 2 in a single throw is [tex] \frac{1}{2} [/tex]
The probability of not getting a 2 in a single throw is [tex] \frac{5}{6} [/tex]
The probability of getting one 2 in any of the 6 tosses is
i.e X=1 is given by ₆C₁
multiplied by the probability of getting one 2 in a toss which is 1/6 and the probability of not getting a 2 in one toss which is
[tex] \frac{5}{6} .\frac{5}{6} .\frac{5}{6} .\frac{5}{6} .\frac{5}{6} [/tex]
so this would combine to ₆C₁ . [tex] [\frac{1}{6}]^{1} .[\frac{5}{6}]^{5} [/tex]
Now similarly for X=2
That is the probability of getting two heads in 6 times is
P(X)= ₆C₂ . [tex] [\frac{1}{6}]^{2} .[\frac{5}{6}]^{4} [/tex]
(there are two chances of getting 1/6 and 4 chances of getting 5/6)
Similarly for X=3
P(X)= ₆C₃ . [tex] [\frac{1}{6}]^{3} .[\frac{5}{6}]^{3} [/tex]
and so on.
Hence this will sum to
P(X=a)= nCa X[tex] p^{a} [/tex] X [tex] x q^{n-a} [/tex]
which is a binomial distribution where
N is the number of tosses
a is the number of desired results or successes
P is the probability of success
q is the probability of failures
Hence this situation follows binomial distribution
