Horwege Electronics Inc. purchases TV picture tubes from four different suppliers. Tyson Wholesale supplies 20 percent of the tubes, Fuji Importers 30 percent, Kirkpatricks 25 percent and Parts Inc. 25 percent. Tyson Wholesale tends to have the best quality, as only 3 percent of its tubes arrive defective. Fuji Importers tubes are 4 percent defective, Kirkpatricks’ 7 percent and Part Inc. are 6.5 percent defective,

1. Example
Horwege Electronics Inc. purchases TV picture tubes from four different suppliers.
TysonWholesale supplies 20 percent of the tubes, Fuji Importers 30 percent,
Kirkpatricks 25 percent and Parts Inc. 25 percent. Tyson Wholesale tends to have
the best quality, as only 3 percent of its tubes arrive defective. Fuji Importers tubes
are 4 percent defective, Kirkpatricks’ 7 percent and Part Inc. are 6.5 percent
defective,
(a) What is the overall percent defective?
(b)A defective picture tube was discovered in the latest shipment. What is the
probability that it came from Tyson Wholesale?
Let:
T = Tyson Wholesale
F = Fuji Importers
K = Kirkpatricks
P = Parts Inc.
Probabilities:
P(T) = 0.20
P(F) = 0.30
P(K) = 0.25
P(P) = 0.25
Conditional Probabilities:
P(Defective / TysonWholesale ) = P(D / T ) = 0.03
P(Defective / Fuji Importers) = P(D / F) = 0.04
P(Defective / Kirkpatricks) = P(D / K) = 0.07
P(Defective / Parts Inc.) = P(D / P) = 0.065

2. (a)
𝑃(𝐷) = 𝑃(𝑇). 𝑃(𝐷
𝑇
⁄ ) + 𝑃(𝐹).𝑃(𝐷
𝐹
⁄ ) + 𝑃(𝐾).𝑃(𝐷
𝐾
⁄ ) + 𝑃(𝑃).𝑃(𝐷
𝑃
⁄ )
𝑃(𝐷) = (0.2× 0.03)+ (0.3 × 0.04) + (0.25× 0.07) + (0.25× 0.065)
𝑃(𝐷) = 0.05175 = 5.175%
(b)
Find 𝑃(𝑇
𝐷
⁄ ) using Bayes'Theorem:
𝑃(𝑇
𝐷
⁄ ) =
𝑃(𝑇).𝑃(𝐷
𝑇
⁄ )
𝑃(𝐷)
𝑃(𝑇
𝐷
⁄ ) =
𝑃(𝑇). 𝑃(𝐷
𝑇
⁄ )
𝑃(𝐷)
𝑃(𝑇
𝐷
⁄ ) =
(0.20× 0.03)
0.05175
=
0.006
0.05175
= 0.1159