The probability that any postcard posted in Portugal on Monday is delivered to the UK within a week is 0.62 - OCR - GCSE Maths - Question 7 - 2018 - Paper 6
Question 7
The probability that any postcard posted in Portugal on Monday is delivered to the UK within a week is 0.62.
The probability that any postcard posted in Portugal o... show full transcript
Worked Solution & Example Answer:The probability that any postcard posted in Portugal on Monday is delivered to the UK within a week is 0.62 - OCR - GCSE Maths - Question 7 - 2018 - Paper 6
Step 1
How many of her postcards can she expect to be delivered within a week?
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Answer
To determine how many postcards Anna can expect to have delivered within a week, we can use the formula for expected value:
E(X)=nimesp
Where:
n is the total number of postcards (15)
p is the probability of delivery within a week (0.62)
Calculating this gives:
E(X)=15imes0.62=9.3
Since the expected number of postcards delivered cannot be fractional, we round this to the nearest whole number. Therefore, Anna can expect 9 postcards to be delivered within a week.
Step 2
Complete the probability tree to show the possible outcomes for the postcards.
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The probability tree can be completed as follows:
Monday's postcard:
Delivered within a week (Probability = 0.62)
Not delivered within a week (Probability = 1 - 0.62 = 0.38)
Friday's postcard:
Delivered within a week (Probability = 0.41)
Not delivered within a week (Probability = 1 - 0.41 = 0.59)
This results in the following probabilities:
Monday's postcard delivered: 0.62
Friday's postcard delivered: 0.41
Friday's postcard not delivered: 0.59
Monday's postcard not delivered: 0.38
Friday's postcard delivered: 0.41
Friday's postcard not delivered: 0.59
Step 3
Calculate the probability that only one of Sergio's postcards is delivered within a week.
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Answer
To calculate the probability that only one of Sergio's postcards is delivered within a week, we can examine the two cases:
Monday's postcard is delivered, and Friday's postcard is not delivered:
Probability = 0.62imes0.59=0.3658
Monday's postcard is not delivered, and Friday's postcard is delivered:
Probability = 0.38imes0.41=0.1558
Now, we sum these two probabilities:
P(extOnlyonedelivered)=0.3658+0.1558=0.5216
Thus, the probability that only one postcard is delivered within a week is 0.5216.