Question

Suppose the number of radios in a household has a binomial distribution with parameters n=19 and p=15%. Find the probability of a household having:

(a) 4 or 18 radios

(b) 16 or fewer radios

(c) 14 or more radios

(d) fewer than 18 radios

(e) more than 16 radios

Answer 1

n = 19

P = 0.15

To find the probability for 1 radio we need to find binomial probability :

P(x) is the probability of x successes out of N trials. = probability of success on a given trial

Here N=19, = 0.15

So for x= 0

We create below table :

P(0) = 0.045599448334723

P(1) = 0.15289226794584

P(2) = 0.24282889614927

P(3) = 0.24282889614927

P(4) = 0.17140863257595

P(5) = 0.090745746657858

P(6) = 0.037365895682647

P(7) = 0.01224596581196

P(8) = 0.0032415791855188

P(9) = 0.00069916413805308

P(10) = 0.00012338190671525

P(11) = 1.7814499900063E-5

P(12) = 2.0958235176544E-6

P(13) = 1.9915065099884E-7

P(14) = 1.5061813941089E-8

P(15) = 8.8598905535816E-10

P(16) = 3.9087752442272E-11

P(17) = 1.2172656469912E-12

P(18) = 2.3867953862572E-14

P(19) = 2.2168378200531E-16

(a) Probability of household having 4 or 18 radios = P(4) + P(18)

=  0.17140863257595 +  2.3867953862572E-14

=   0.17140863257

(b) 16 or fewer

P(X≤16) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15)+P(16)

P(X≤16) = 0.99999999999876

(c) 14 or more radios

P(X≥14) = P(14)+P(15)+P(16)+P(17)+P(18)+P(19)

P(X≥14) = 1.5988132104174E-8

(d) Fewer than 18

P(X<18) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)

P(X<18) Probability of less than 18 successes: 0.99999999999998

(e) more than 16

P(x>16) = P(17) + P(18)+P(19)

P(x>16) = 1.2413552846357E-12