Question

Exercise 4. Find the probability to get: (a) the number "3" for the 4th time at the 7th toss of a fair die. (b) two heads for the first time at the 10th toss of two fair coins. (c) the red ball for the 2nd time in the 5th selection when selecting randomly a ball with replacement from a box containing 4 red and 6 blue balls.

Answer 1

a) Number 3 has occurred 4th time at the 7th toss of a fair die. This means that in the previous six tosses three times 3 has appeared. We know the probability of a 3 at the 7th toss = 0.5.

Let X denote the number of 3's in the first six tosses then X follows a binomial distribution with n=6 and p=1/6. Thus the probability of getting three times 3 in first six throws is

$\small P(X=3)=\binom63* \frac{1}{6^3}*( \frac{5}{6})^3=0.05358$. Hence the probability of getting three 4th time at the 7th toss of a fair die = 0.0538*0.5 = 0.0269

b) If success is considered to be getting two heads in a throw of two fair coins, then the probability of getting success is 1/4. Thus is Y denote the number of throws required to get the success then Y follows a geometric distribution with p=0.25. The pmf of Y is given by:

$\small P(Y=y)=0.25*(0.75)^{y-1}\ for\ y=0,1,2,....$

Thus the probability of getting the success at the 10th toss is:

$\small P(Y=y)=0.25*(0.75)^9=0.0187$

c) This problem is similar to part (a). Here color red is obtained for the 2nd time in 5th chance. Thus in the first 4 chances red was obtained only one time.

P(getting red in 5th chance) = 0.4.

Now let Z denote the number of red balls obtained in the first 4 chances then X follows a binomial distribution with n=4 and p=0.4. then probability of getting 1 red in first 4 chances = P(Y = 1) = 4*0.4*0.66= 0.3456.

Hence P( red is obtained for the 2nd time in 5th chance) = 0.3456*0.4 = 0.13284