Question

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

Answer 1

Hint: 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\% $
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\% $ (ten percent)
So first we derive this fully we get; \[10\% = \dfrac{{10}}{{100}} = \dfrac{1}{{10}}\](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 $1 - \dfrac{1}{{10}} = \dfrac{9}{{10}}$(diseases not fatal)
Since the total number of patients is $5$ and the patients that exactly or only need to die is $3$
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 ${}^5{c_3}$ and $\dfrac{1}{{10}}$ is the diseases fatal and $\dfrac{9}{{10}}$ is the diseases that not fatal;
Hence the required probability is ${}^5{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3}$ (since only three needs to die is the restriction)
Further solving we get; \[{}^5{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} = \dfrac{{5 \times 4}}{2} \times \dfrac{1}{{{{10}^3}}} \times \dfrac{9}{{{{10}^3}}}\] (thus solving by cancellation)
${}^5{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} = 729 \times {10^{ - 5}}$ (division power will upper comes with the negative)
Hence option $2)729 \times {10^{ - 5}}$ is correct.

Note: Since the combination formula is ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$ 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 $3)729 \times {10^{ - 6}}$ is wrong
And after solving the probability we only get $729 \times {10^{ - 5}}$and thus $1)8748 \times {10^{ - 5}}$, $4)41 \times {10^{ - 6}}$ are also wrong with no comparison.
General probability if $P = \dfrac{F}{T}$ favorable events divide the total events.