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Question: The records of a hospital show that 10% of the cases of a certain disease are fatal. If 6 patients a...

The records of a hospital show that 10% of the cases of a certain disease are fatal. If 6 patients are suffering from the disease, then the probability that only three will die is

A

1458×10−51458 \times 10 ^ { - 5 }

B

1458×10−61458 \times 10 ^ { - 6 }

C

41×10−641 \times 10 ^ { - 6 }

D

8748×10−58748 \times 10 ^ { - 5 }

Answer

1458×10−51458 \times 10 ^ { - 5 }

Explanation

Solution

Probability of disease is fatal = p=10%p = 10 \%

p=10100=110,q=910p = \frac { 10 } { 100 } = \frac { 1 } { 10 } , q = \frac { 9 } { 10 }

Number of patients = 6, Number of die cases = 3

∴\thereforeProbability that 3 will die

=6C3(110)3(910)3=1458×10−5= { } ^ { 6 } C _ { 3 } \left( \frac { 1 } { 10 } \right) ^ { 3 } \left( \frac { 9 } { 10 } \right) ^ { 3 } = 1458 \times 10 ^ { - 5 } .