Question

Find probability of throwing at most one 6 in 6 throws of a single day

Answer 1

Let X be the number of times we get 6 in six throws of a die. P (getting a 6) =p=16→q=1−p=56

Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, P(X=x)=nCx.px.qn–x where x=0,1,2,...,n and (q=1–p)

P (throwing at least 2 sixes) =P(X≤2)=P(X=0)+P(X=1)+P(X=2)

P(X=0)=6C0.160.566–0=(56)6

P(X=1)=6C1.161.566–1 =6×16×(56)5=(56)5

P(X=2)=6C2.162566–2=15162(56)4

Therefore P(X≤2)=(56)6 +(56)5 +15162(56)4

P(X≤2)=(56)4((56)2+56+1536)

P(X≤2)=(56)425+30+1536

P(X≤2)=(56)43518

Answer 2

Answer:

1/6

Step-by-step explanation:

there are 6 possible outcomes on 1 throw and there will be 36 possible outcomes on 6 throws. So six can fall 6 times.

6/36 = 1/6