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Question

Mathematics Question on Probability

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Answer

s={1,2,3,4,5,6}
⇒\Rightarrow n(S)=6
Let A represents the favorable event i.e.,6
A={6]⇒\Rightarrow n(A)=1

P=n(A)n(S)=16\frac{n(A)}{n(S)}=\frac{1}{6} and q=1-p=1−16=561-\frac{1}{6}=\frac{5}{6}

n=6,r=0,1,2 and P(X=r)=nCrprqn−r^nC_rp^rq^{n-r}
P(X=0)=(56)6\bigg(\frac{5}{6}\bigg)^6
P(X=1)=6C1(16)(56)5=(56)5^6C_1\bigg(\frac{1}{6}\bigg)\bigg(\frac{5}{6}\bigg)^5=\bigg(\frac{5}{6}\bigg)^5

P(X=2)=6C2(16)2(56)4=15.(16)2(56)4^6C_2\bigg(\frac{1}{6}\bigg)^2\bigg(\frac{5}{6}\bigg)^4=15.\bigg(\frac{1}{6}\bigg)^2\bigg(\frac{5}{6}\bigg)^4

P(at most 2 success)=P(X=0)+P(X=1)+P(X=2)

=(56)6+(56)5+15.(16)2+(56)4\bigg(\frac{5}{6}\bigg)^6+\bigg(\frac{5}{6}\bigg)^5+15.\bigg(\frac{1}{6}\bigg)^2+\bigg(\frac{5}{6}\bigg)^4

=(56)4(2536)+(56)+(1536)=(56)4(7036)=(56)4(3518)\bigg(\frac{5}{6}\bigg)^4\bigg(\frac{25}{36}\bigg)+\bigg(\frac{5}{6}\bigg)+\bigg(\frac{15}{36}\bigg)=\bigg(\frac{5}{6}\bigg)^4\bigg(\frac{70}{36}\bigg)=\bigg(\frac{5}{6}\bigg)^4\bigg(\frac{35}{18}\bigg)