http://www.ck12.org Chapter 3. Introduction to Discrete Random Variables
P(X=a) =nCa×pa×q(n−a)
P(X= 2 ) = 15 C 2 ×p^2 ×q(^15 −^2 )
P(X= 2 ) = 15 C 2 ×( 0. 167 )^2 ×( 0. 833 )(^15 −^2 )
P(X= 2 ) = 105 × 0. 0279 × 0. 0930
P(X= 2 ) = 0. 272Therefore, the probability of seeing a 2 exactly twice when a die is rolled 15 times is 27.2%.
It should be noted here that the examples involve binomial experiments. Again, we will be learning more about
binomial experiments and distributions in later Concepts. For now, we can visualize abinomial distribution
experiment as one that has a fixed number of trials, with each trial being independent of the others. In other words,
rolling a die twice to see if a 2 appears is a binomial experiment, because there is a fixed number of trials (2), and each
roll is independent of the others. Also, for binomial experiments, there are only 2 possible outcomes (a successful
event and a non-successful event). For our rolling of the die, a successful event is seeing a 2, and a non-successful
event is not seeing a 2.
Guided Practice
A pair of fair dice is rolled 10 times. LetXbe the number of rolls in which we see at least one 2.
(a) What is the probability of seeing at least one 2 in any one roll of the pair of dice?
(b) What is the probability that in exactly half of the 10 rolls, we see at least one 2?
Answer:
If we look at the chart below, we can see the number of times a 2 shows up when rolling 2 dice.
(a) The probability of seeing at least one 2 in any one roll of the pair of dice is:
P(X) =
11
36
= 0. 306
(b) The probability of seeing at least one 2 in exactly 5 of the 10 rolls is calculated as follows:
p= 0. 306
q= 1 − 0. 306 = 0. 694
n= 10
a= 5