The probability that a shopper will buy a PS3 is 0.40. The probab

siroticuvj

siroticuvj

Answered question

2021-11-17

The probability that a shopper will buy a PS3 is 0.40.
The probability that a shopper will buy an X-Box 360 is 0.54.
The probability that a shopper will buy a Nintendo DS is 0.30.
The probability that a shopper will buy both a PS3 and an X-Box 360 is 0.13. The probability that a shopper will buy both a PS-3 and a DS is 0.12.
The probability that a shopper will buy both an X-Box 360 and a DS is 0.07. There is no chance that a shopper will buy all three systems.
Find the probability that a random shopper will:
a) Buy only a PS-3
b) Buy exactly one type of system
c) Buy exactly two types of systems d) Not buy any of the three systems

Answer & Explanation

Mary Ramirez

Mary Ramirez

Beginner2021-11-18Added 19 answers

Step 1
Let;
PS=PS3
XB=XB 360
ND= Nintendo DS
Given that;
P(PS)=0.40
P(XB)=0.54
P(ND)=0.30
P(PSXB)=0.13
P(PSND)=0.12
P(XBND)=0.07
P(PSXBND)=0
Step 2
a) Probability that a random shopper will buy only a PS-3.
P(only a PS3)
=P(PS)P(PSXB)P(PSND)
=0.400.130.12
=0.15
b) Probability that a random shopper will buy exactly one type of system.
P(PSXBND)=P(PS)+P(XB)+P(ND)P(PSXB)
P(PSND)P(XBND)
+P(PSXBND)
=0.40+0.54+0.300.130.120.07+0
=0.92
c) Probability that a random shopper will buy exactly two types systems.
P(exactly two types of systems)
=P(PSXB)+P(PSND)+P(XBND)
=0.13+0.12+0.07
=0.32
d) Probability that a random shopper will not buy any of the three systems.
P(PSCXBCNDC)
=P(PSXBND)C [De Morgan Law]
=1P(PSXBND)
=10.92
=0.08

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