Examples of nursing interventions and rationales

Binomial D
istribution
After studying the random variables and discrete probability distributions, we need to look a little more
closely at a special type of discrete distribution, one that is closely related to
the example we used
earlier about the
number of boys when you have 4 kids. What characterize these types of distributions is
that they can all be see
n as repeated
coin flips
–
you can call the two outcomes boys/girls, heads/tails,
or whatever. But the mathematics is really the same.
Binomial
Random Distribution based on a Fair Coin
Suppose we have a fair coin (so the heads
-
on probability is 0.5), and we flip it 3 times. If we let the
random variable X represent the number of heads in the 3 tosses, then clearly, X is a discrete random
variabl
e, and can take values ranging from 0 to 3. Moreover, we can represent the probability
distribution of X in the following table:
x
P(x)
0
1
2
3
What will be the probability of
0
X

? We can do it in one of two ways: since the coin is fair, the
sample space has 8 equally likely outcomes {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. So the chance of
0 heads is just the probability of getting all tails, i.e.
(
0) 1 / 8
PX

. Alte
rnatively, we can also use
independence of the 3 coin tosses, and break down the probability using the multiplication rule, applied
to the independent events:
3
1
(
0)
(
)
( )
( )
( ) ( )
1 / 8
2
PXPHHHPHPHPH
 



 
. Using the
same reasoning, it's also easy to determine that
(
3) 1 / 8
PX

. So our probability distribution now
becomes:
x
P(x)
0
1/8
1
2
3
1/8
The probability
(
1)
PX

is slightly more complicated. Using the classical probability approach, the
event that "there is 1 head in 3 tosses" includes 3 separate outcomes: {HTT, THT, TTH}. So the
probability of getting 1 head is
(
1) 3 / 8
PX

.
But there is another w
ay that uses the additional rule and multiplication rule together that you should
know: since "1 head" is technically a union of the three outcomes: HTT, THT, TTH, and they are mutually
exclusive;
we may use the addition rule applied to the mutually exclus
ive events:
(
1)
(HTT or THT or TTH)
PXP

(
)
(
)
(
)
PHTTPTHTPTTH



1 1 1
1 1 1
1 1 1
2 2 2
2 2 2
2 2 2
        
You probably noticed that the three probabilities being added are all the same, so we can shorten it as:
3
13
(
1) 3
28
PX

  



So the factor of 3 comes from the use of the addition rule, and the
3
(1 / 2)
is a result of applying the
multiplication rule. Using the same type of analysis, we can also determine that
(
2) 3 / 8
PX

.
Hence the completed prob
ability distribution is the following:
x
P(x)
0
1/8