# S2 questionsWatch

This discussion is closed.
#1
(1) i can do the question but the end part states calculate the least number of x in order of getting at least one suceess exceeds 0.99. How do i go about this?? (Heinemann S2 ex 1D question 6 last bit)

(2) Poisson random variable with parameter 2. Factory observed for 100 weeks. Determine expected number of weeks in which 5 or more accidents occur.

(3) an archer fires arrows at a target and for each arrow, independently of all others, the prob that hits bulles eye is 1/8 Find the smallest value of m such that prob archer hits bulls eye with at least m arrows is greater than 0.5
0
13 years ago
#2
(2)
100*P(X >= 5)
= 100*(1 - e^(-2)(1 + 2 + 2^2/2! + 2^3/3! + 2^4/4!))
= 5.265

(3)
P(hits with at least one arrow)
= 1 - P(misses with all arrows)
= 1 - (7/8)^m

m | P(hits with at least one arrow)
1 | 0.125
2 | 0.234
3 | 0.330
4 | 0.414
5 | 0.487
6 | 0.551

0
#3
for numer 2 the answer in back says 0.2707

for 3 it says m=3

????????????
0
13 years ago
#4
(Original post by Sharma)
for numer 2 the answer in back says 0.2707

for 3 it says m=3

????????????
I think you have mistyped Question 3. How many arrows does the archer shoot?
0
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