Question

The probability that a patient recovers from a rare blood disease is 0.4 . If 15 people are known to have contracted the disease what is the probability that: a. At least 10 survive b. Between 3 and 8 survive c. Exactly 6 survive

Answer 1

Hello Sir/ Mam

Given is the case of binomial distribution where there are 2 possible choices, whether the patient recovers from it or not. Hence,

Using this formula,

p	40.00%
n	15
x
x	P(x)
0	0.000470
1	0.004702
2	0.021942
3	0.063388
4	0.126776
5	0.185938
6	0.206598
7	0.177084
8	0.118056
9	0.061214
10	0.024486
11	0.007420
12	0.001649
13	0.000254
14	0.000024
15	0.000001

Hence,

(a)

P(Atleast 10 survivors) = 0.0338 = 3.38%

(b)

P(Between 3 and 8 survivors) = 0.6964

(c)

P(Exactly 6 survivors) = 0.2066

I hope this solves your doubt.

Do give a thumbs up if you find this helpful.