Question

Refer to the table below. Given that 2 of the 127 subjects are randomly​ selected, complete parts​ (a) and​ (b).

    Rh +   Rh -
O   43   12
A   31   10
B   11   4
AB   13   3

a. Assume that the selections are made with replacement. What is the probability that the 2 selected subjects are both group AB and type Rh +?

b. Assume the selections are made without replacement. What is the probability that the 2 selected subjects are both group AB and type Rh+?

Answer 1

Solution:

	Rh+	Rh-	total
O	43	12	55
A	31	10	41
B	11	4	15
AB	13	3	16
total	98	29	127

( a )

   selections are made with replacement

so , selection is independent

n = 2

p = 0.1023

Formula:

P_{(k out of n )}= n!*p^k * q^n-k / k! *(n - k)!

P( x = 2 ) = 2!*0.1023² * 0.8977^2-2 / 2! *(2 - 2)!

= 0.0105

( b )

selection is made without replacement

so,

the total number of subjects and the subjects with both type AB and Rh+ changes after the first selection

total number of subjects = 127

number of subjects who are both group AB and type Rh+ = 13

probability of selecting first subject who is both group AB and Rh+ = 13 / 127 = 0.1023

after the fist selection,

number of remaining total subjects = 127 - 1 = 126

number of remaining subject who is both group AB and Rh+ = 13- 1 = 12

probability of selecting second subject who is both group AB and Rh+ = 12 / 126 = 0.0952

now,

probability of selecting two subjects who are both group AB and Rh+ = 0.1023 * 0.0952 = 0.0097