Question

#1 A coin will be tossed 10 times. Find the chance that there will be exactly 2 heads among the first 5 tosses, and exactly 4 heads among the last 5 tosses.

#2 An 11-digit number is randomly chosen by drawing 11 times from a box that has one ticket for each of the numbers 0 to 9 and writing down the numbers on the tickets in the order in which they are drawn. Find the chance that exactly 4 of the digits in the number chosen are five.

Answer 1

Question 1:

The number of heads in the first 5 tosses could be modelled as:

Xi ~ Binn = 5. P= 0.5) (

Similarly for last 5 tosses, we have the distribution as:

X2 ~ Binn = 5. P-0.5) (

Also as the first 5 tosses are independent of the last 5 tosses, the required probability here is computed as:

$P(X_1 = 2, X_2 = 4) = P(X_1 = 2)P(X_2 = 4) = \binom{5}{2}0.5^5*\binom{5}{4}0.5^5 = 0.0488$

Therefore 0.0488 is the required probability here.

Question 2:

Here as each of the 11 digits is equally likely to be any digit from 0 to 9, therefore for each draw there is a probability of 0.1 to get a digit five. The number of fives in 11 draws could be modelled here as:

$Y \sim Bin(n = 11, p = 0.1)$

Therefore now the required probability is computed here as:

$P(Y = 4) = \binom{11}{4}0.1^4(1- 0.1)^7 = 0.0158$

Therefore 0.0158 is the required probability here.