Answer to Question #126302 in Statistics and Probability for Anelisa

Question

1 Determine the probability of a sum of 11 or 9 appears in a single toss of a pair of fair dice
2 At least one tail appears in two tosses of a fair coin
3 In an experiment a coin is tossed 4 times.Find the probability of getting at least 2 heads and 1tail.

Answer 1

1.Let A be the event that "a sum of 11 or 9 appears in a single toss of a pair of fair dice"

P(A)=m/n

There are six ways to implement event A:

11=5+6

11=6+5

9=3+6

9=4+5

9=5+4

9=6+3

So m=6.

There are 6²=36 ways to roll two dice:

(i,j), i=1,2,3,4,5,6, j=1,2,3,4,5,6

So n=36

P(A)=6/36=1/6

2.Let B be the event that "at least one tail appears in two tosses of a fair coin"

P(B)=m/n

There are 4 ways in two tosses of a fair coin:

(H,H),(H,T),(T,H),(T,T)

So n=4

There are 3 ways implement event B:

(H,T),(T,H),(T,T).

So m=3

P(B)=3/4.

3.Let C be the event that "at least 2 heads and 1 tail appear in an experiment a coin is tossed 4 times"

P(C)=m/n

There are 2⁴=16 ways in an experiment a coin is tossed 4 times:

(H,H,H,H),(H,H,H,T),(H,H,T,H),(H,H,T,T),

(H,T,H,H),(H,T,H,T),(H,T,T,H),(H,T,T,T),

(T,H,H,H),(T,H,H,T),(T,H,T,H),(T,H,T,T),

(T,T,H,H),(T,T,H,T),(T,T,T,H),(T,T,T,T).

So n=16.

There ₄C₂=4!/2!/2!=6 ways 2 heads and 2 tail appear in an experiment:

(H,H,T,T),(H,T,H,T),(H,T,T,H),(T,H,H,T),(T,H,T,H),(T,T,H,H).

There ₄C₁=4!/1!/3!=4 ways 3 heads and 1 tail appear in an experiment:

(H,H,H,T),(H,H,T,H),(H,T,H,H),(T,H,H,H).

So m=6+4=10.

P(C)=10/16=5/8.

Answer:

1.The probability of a sum of 11 or 9 appears in a single toss of a pair of fair dice is equal to 1/6.

2.The probability of event that at least one tail appears in two tosses of a fair coin is equal to 3/4.

3.The probability of getting at least 2 heads and 1 tail is equal to 5/8.