Anne, Edward, and Isaac are the three editors of the Perplexus Weekly newspaper. This week, Anne spotted 300 errors, Edward spotted 100 errors, and Isaac spotted 200 errors. Altogether, they spotted 404 errors. Assuming that each error is equally easy to spot, about how many errors did they miss?
(In reply to
Answer by K Sengupta)
In the given problem, let n = total number of errors.
Now, the respective number of errors spotted by each of Anne, Edward and Isaac are 300, 100 and 200.
Let P, Q and R denote the respecive probability that Anne, Edward and Isaac can spot a given error.
Then, (P, Q, R) = (300/n, 100/n, 200/n) ....(i)
Let S denote the event that a given error is not spotted by all three editors.
Since the three editors altogether spotted 404 errors, it follows that:S = 404/n ......(ii)
Thus, 1-P, 1-Q, 1-R, 1-S) = (1- 300/n,1- 100/n, 1- 200/n, 1- 404/n)
Now, we know that if two or more events are independent, then their joint probability is the product of the prior individual probabilities of each event.
Thus, we must have:
1- 404/n = (1- 300/n) (1- 100/n)(1- 200/n)
-> n^2(n-404) = (n-300)(n-100)(n-200)
-> 196*n^2 - 110000n + 6000000 = 0
-> 49*n^2 - 27500n + 1500000 = 0
-> 49*n^2 - 24500n -3000n + 1500000 = 0
-> 49n(n-500) - 3000(n-500) = 0
-> (n-50)(49n- 3000) = 0
-> n = 500,Or 3000/49
obviously, the total number of missed errors must be an integer, and accordingly n = 500
Consequently, the total number of errors missed by the three editors is 500.
Puzzle Solution
