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Amiraneli [1.4K]
3 years ago
9

the probability of a train arriving on time and leaving on time is 0.8. the probability of the same train arriving on time is 0.

84. the probability of of this train leaving on time is 0.86. given the train arrived on time, what is the probability it will leave on time?
Mathematics
2 answers:
horrorfan [7]3 years ago
8 0
Let the event 'arrive on time' be AO.
Let the event 'leave on time' be LO.
P(AO\cap LO)=0.8
P(AO) = 0.84
P(LO) = 0.86
P(LO|AO)=\frac{P(AO\cap LO)}{P(AO)}=\frac{0.8}{0.84}=0.95
mina [271]3 years ago
4 0

Answer: Our required probability is 0.952.

Step-by-step explanation:

Since we have given that

Probability that train arrive on time and leaving on time = 0.8

Probability that train arrive on time = 0.84

Probability that train leave on time = 0.86

We need to find that probability that it will on time given that train arrived on time.

We will use "Conditional probability":

P(\text{Leave on time}|\text{arrive on time})=\dfrac{P(\text{Arrive}\cap \text{Leave})}{P(Arrive)}=\dfrac{0.8}{0.84}=0.952

Hence, our required probability is 0.952.

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Mice21 [21]

Answer:

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0.023........3 decimal point

    2345

×       23               1.    first we change it whole number for multiply.

····················          2.count the decimal point.here  6 decimal points are there.

+ 7 0 3  5          3.we got the answer 53935.here only 5 digits are there.

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····························5 . final answer is   0.053935

 5  3  9  3   5

···························

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Step-by-step explanation:

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