a) An urn contains pink and green flowers. Two flowers are chosen with

kiki195ms

kiki195ms

Answered question

2021-11-15

a) An urn contains pink and green flowers. Two flowers are chosen without replacement. The probability of selecting a ping flower and then a green flower is 0.73, and the probability of selecting a pink flower on the first draw is 0.24. What is the probability of selecting a green flower on the second draw, given that the first flower drawn was pink?
b) In a bolt factory, three machines Z, F and W makes 10%, 60% and 30% respectively of the products. It is known from the past experience that 4%, 7% and 9% of the products by each machine respectively are defective. Now suppose that a finished product is randomly selected. If the selected bolt is defective then what is the probability that the bolt was produced by Z machine.

Answer & Explanation

Mike Henson

Mike Henson

Beginner2021-11-16Added 11 answers

Step 1 Introduction:
Probability: Probability is chance factor it will occur at every outcome of a random experiment.The probability is defined as the ratio of favorable outcomes to the total number of possible outcomes and it is denoted as
p= favorable outcomes/possible outcomes
probability value will always lie between 0 and 1
Conditional Probability: The probability of an event occurring given that another event has already occurred is called
a conditional probability
The formula for the conditional probability
P(AB)=PAandBP(B)
a. we have p(pink and green) is 0.73 and p(pink) is 0.24
p(greenπnk)=pπnkandgreenp(πnk)
=0.730.24
=3.04
the probability value does not exceed 1
Therefore, The given data is inadequate because, the probability values will always lie between 0 and 1.
Step 2 Calculation
b. we have P(Z)=10100=0.1
P(F)=60100=0.6
P(W)=30100=0.3
P(DZ)=4100=0.04
P(DF)=7100=0.07
P(DW)=9100=0.09
where D denotes the defective products
Now P(D)=P(Z)P(DZ)+P(F)P(DF)+P(W)P(DW)
=0.10.04+0.60.07+0.30.09
=0.004+0.042+0.027
=0.073
The probability that the defective bulb produced by machine Z
P(ZD)=P(Z)PDZP(Z)P(DZ)+P(F)P(DF)+P(W)P(DW)
=0.0040.073
=0.05
Therefore,The probability that the defective bulb produced by machine Z is 0.05

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