Answer to Question #244673 in Statistics and Probability for Nana

Question

(a) A fair coin is tossed until a head is obtained. What is the probability that the number of tosses required is an odd number?

(b) A fair die is thrown seven times. Find the probability that the outcome contains one ‘1’, two ‘2’, three ‘3’, but no ‘4’?

I would like to know how to calculate these questions, thanks!

Answer 1

Solution:

(a):

S={H,TH,TTH,TTTH,...}

We want the number of tosses to be odd.

P(H) = 1/2

P(TTH) = $\dfrac1{2^3}$

P(TTTTH) = $\dfrac1{2^5}$

Required probability$=\dfrac12+\dfrac1{2^3}+\dfrac1{2^5}+...$

$=\dfrac{\frac12}{1-\frac14}=\dfrac{2}{3}$

(b):

Clearly the sample will be of size 6⁷.

Now, we will count the number of ways one can arrange one 1, two 2s, three 3s :

So, basically we have 6 symbols of which two are same and the another three are same and one is single. Here the counting goes by the following 6!/(1! 2! 3!) =60

Now we have only two symbols 5 and 6 to choose from, as we can't choose 4 and 1, 2 and 3 are done.

so we can make that choosing in 2 ways and we can put those two in 7 places as we can obtain 5 or 6 at any point of throw.

So, putting 5 or 6 in between the throws can be done in 2*7=14 ways.

So, total number of counting in our favour will be 14*60.

So the required probability will be 14*60 / (6⁷) $=\dfrac{35}{11664}$