Question

The record of a hospital shows that $10\% $ of the cases of a certain disease are fatal. If $6$ are suffering from the disease, then the probability that only $3$ will die is
$1)1458 \times {10^{ - 5}}$
$2)1458 \times {10^{ - 6}}$
$3)41 \times {10^{ - 6}}$
$4)8748 \times {10^{ - 5}}$

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Since the hospital records show that the cases that particular disease is fatal are at a percentage $10\% $
Also, there are a total of six 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 $6$and the patients that exactly or only need to die is $3$
Hence here we use the method of combination that is the number of ways to find the required resultant.
The first total of six patients and three need to die combination is ${}^6{c_3}$ and $\dfrac{1}{{10}}$is the diseases fatal and $\dfrac{9}{{10}}$is the diseases that not fatal;
Hence the required probability is ${}^6{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3}$(since only three needs to die is the restriction)
Further solving we get; ${}^6{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} = \dfrac{{6 \times 5 \times 4}}{{2 \times 3}} \times \dfrac{1}{{{{10}^3}}} \times \dfrac{9}{{{{10}^3}}}$(thus solving by cancellation)
${}^6{c_3} \times {(\dfrac{1}{{10}})^3} \times {(\dfrac{9}{{10}})^3} = 1458 \times {10^{ - 5}}$(division power will upper comes with the negative)
Hence option $1)1458 \times {10^{ - 5}}$ is correct.

Note: Since the combination formula is ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Also, all other options are wrong where the power ten is minus five thus $2)1458 \times {10^{ - 6}}$is wrong
And after solving the probability we only get $1458 \times {10^{ - 5}}$and thus $3)41 \times {10^{ - 6}}$,$4)8748 \times {10^{ - 5}}$ are also wrong with no comparison.