求數學高手入內指教probability問題, 謝謝各位!!

simonkung2008

2012-11-15 9:51 pm

請以列明計算步驟及答案( answer in English please ), 謝謝指教

Question:
A fast food shop found that 40% of all customers use mustard, 70% use ketchup, and 20% use both, when eating a hotdog.

Q1: What is the probability that a particular customer will use exactly one or two sauces ( mustard, ketchup ) when eating a hotdog?

Q2: What is the probability that a particular customer will use neither mustard nor ketchup when eating a hotdog?

Q3: A customer just said that he didn't want ketchup, what is the probability that he will use mustard?

Q4: What is the probability that among the next four customers, none of them use mustard? ( Assume they make their orders independently )

回答 (1)

賣女孩的火柴

2012-11-15 10:40 pm

✔ 最佳答案

Denotation :
M : a customer uses mustard
M' : a customer does not use mustard
K : a customer uses ketchup
K' : a customer does not use ketchup

Given that :
P(M) = 40% =0.4
P(M') = 1 - 40% = 0.6
P(K) = 70% = 0.7
P(K') = 1 - 70% = 0.3
P(M∩K) = 0.2

Q.1
The required probability
= P(MUK)
= P(M) + P(K) - P(M∩K)
= 0.4 + 0.7 - 0.2
= 0.9

Q.2
The required probability
= P(M'∩K')
= 1 - P(MUK)
= 1 - 0.9
= 0.1

Q.3
P(M∩K')
= 0.4 - 0.2
= 0.2

The required probability
= P(M|K')
= P(M∩K')/P(K')
= 0.2/0.3
= 2/3

Q.4
The required probability
= [P(M')]⁴
= (0.6)⁴
= 0.1296

參考: 賣女孩的火柴

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