Question

Consider rolling a fair die twice and tossing a fair coin nineteen times. Assume that all the tosses and rolls are independent.

The chance that the total number of heads in all the coin tosses equals 9 is(Q)_____ , and the chance that the total number of spots showing in all the die rolls equals 9 is(Q)__________ The number of heads in all the tosses of the coin plus the total number of times the die lands with an even number of spots showing on top (Q)______(Choose A~E)

a. has a Binomial distribution with n=31 and p=50%
b. does not have a Binomial distribution
c. has a Binomial distribution with n=21 and p=50%
d. has a Binomial distribution with n=21 and p=1/6
e. has a Binomial distribution with n=31 and p=1/6

Answer 1

Hence the correct option is option c has a Binomial distribution with n=21 and p=50%.

Explanation:

1)

A coin is tossed 19 times,

P(Head)=0.5

P(Tail)=0.5

We have to find the probability of a total number of heads in all the coin tosses equals 9.

This can be solved using the binomial distribution. For binomial distribution,

P(X=x)=C(n,x)px(1-p)n-x

where n is the number of trials, x is the number of successes, p is the probability of success, C(n,x) is a number of ways of choosing x from n.

P(X=9)=C(19,9)(0.5)9(0.5)10

P(X=9)=0.1762

2)

A fair die is rolled twice.

Total number of outcomes=36

Possibilities of getting sum as 9

S9={(3,6),(4,5)(5,4),(6,3)}

The total number of spots showing in all the die rolls equals 9 =4/36=0.1111

3)

The event of getting a good number of spots on a die roll is actually no different from the event of heads on a coin toss since the probability of a good number of spots is 3/6 = 1/2, which is additionally the probability of heads. the entire number of heads altogether the tosses of the coin plus the entire number of times the die lands with a good number of spots has an equivalent distribution because the total number of heads in 19+2= 21 tosses of the coin. The distribution is binomial with n=21 and p=50%.