Question

You roll a number cube and flip a coin. Which probability does not belong with The other three

Answer 1

Given:

We roll a number cube and flip a coin.

The sample space is

$S=\mleft\lbrace(1,T),(2,T),(3,T),(4,T),(5,T),(6,T),(1,H),(2,H),(3,H),(4,H),(5,H),(6,TH)\mright\rbrace$$n(S)=12$

1)

Let A be the event getting the number less than 2 and heads.

$A=\lbrack(1,H)\}$$n(A)=1$

The probability is

$P(A)=\frac{n(A)}{n(S)}=\frac{1}{12}$

2)

Let B be the event getting the number less than 2 and tails.

$B=\lbrack(1,T)\}$$n(B)=1$

The probability is

$P(B)=\frac{n(B)}{n(S)}=\frac{1}{12}$

3)

Let C be the event getting the number greater than 2 and tails.

$C=\mleft\lbrace(3,T),(4,T),(5,T),(6,T)\mright\rbrace$$n(C)=4$

The probability is

$P(C)=\frac{n(C)}{n(S)}=\frac{4}{12}=\frac{1}{3}$

4)

Let D be the event getting the number greater than 5 and heads.

$D=\mleft\lbrace(6,H\mright)\}$

$n(D)=1$

The probability is

$P(D)=\frac{n(D)}{n(S)}=\frac{1}{12}$

Hence we get

P(less than 2 and tails)=1/12

P(less than 2 and heads)=1/12

P(greater than 2 and tails)=1/3

P(greater than 5 and heads)=1/12

P(greater than 2 and tails) is not belonging to others.