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Question: If the record of a hospital shows that \(10\% \) of the cases of a certain disease are fatal. If \(5...

If the record of a hospital shows that 10%10\% of the cases of a certain disease are fatal. If 55 are suffering from the disease, then the probability that only 33 will die is
1)8748×10−51)8748 \times {10^{ - 5}}
2)729×10−52)729 \times {10^{ - 5}}
3)729×10−63)729 \times {10^{ - 6}}
4)41×10−64)41 \times {10^{ - 6}}

Explanation

Solution

First, the probability is the concept of a favorable event divided by the total outcome events. Since the hospital records show that the cases that particular disease is fatal are at a percentage 10%10\%
Also, there are a total of five patients suffering from the disease and we will need to find the probability of the patient dying at exactly three only.

Complete step-by-step solution:
Since from the given problem that the probability of the diseases is fatal is 10%10\% (ten percent)
So first we derive this fully we get; 10%=10100=11010\% = \dfrac{{10}}{{100}} = \dfrac{1}{{10}}(which is the probability that diseases if fatal)
And now the probability that the disease is not fatal is hundred percent subtracts the ten percent (disease fatal) hence we get 1−110=9101 - \dfrac{1}{{10}} = \dfrac{9}{{10}}(diseases not fatal)
Since the total number of patients is 55 and the patients that exactly or only need to die is 33
Hence use the method of combination that is the number of ways to find the required resultant.
The first total of five patients and three need to die combination is 5c3{}^5{c_3} and 110\dfrac{1}{{10}} is the diseases fatal and 910\dfrac{9}{{10}} is the diseases that not fatal;
Hence the required probability is 5c3×(110)3×(910)3{}^5{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} (since only three needs to die is the restriction)
Further solving we get; 5c3×(110)3×(910)3=5×42×1103×9103{}^5{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} = \dfrac{{5 \times 4}}{2} \times \dfrac{1}{{{{10}^3}}} \times \dfrac{9}{{{{10}^3}}} (thus solving by cancellation)
5c3×(110)3×(910)3=729×10−5{}^5{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} = 729 \times {10^{ - 5}} (division power will upper comes with the negative)
Hence option 2)729×10−52)729 \times {10^{ - 5}} is correct.

Note: Since the combination formula is ncr=n!r!(n−r)!{}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}} where total given numbers are n-objects and possible outcome number of ways is r-objects.
Also, all other options are wrong where the power ten is minus five thus 3)729×10−63)729 \times {10^{ - 6}} is wrong
And after solving the probability we only get 729×10−5729 \times {10^{ - 5}}and thus 1)8748×10−51)8748 \times {10^{ - 5}}, 4)41×10−64)41 \times {10^{ - 6}} are also wrong with no comparison.
General probability if P=FTP = \dfrac{F}{T} favorable events divide the total events.