Example:
A burglar breaks a stained-glass window to enter a property. 20% of the window glass is red and all the broken glass fragments are well mixed together on the floor. When he is arrested, 10 fragments of glass are found embedded in his new shoes. Assuming all this glass comes from the broken window and that all colours fragment and adhere equally, calculate the probability of finding
- No red fragments, (b) one red fragment and (c) five red fragments.
Solution
Given that a glass fragment is red with probability 0.2 or that it is not red with probability 1-0.2 = 0.8. There are 10 fragments from which we sample m and need to calculate probabilities for specified values of m, all being red fragments.
- For this case m=0 and by substitution into the binomial probability formula, we obtain:
10 ḷ
P(0) = —————- (0.2)0 (1-0.2)10-0 = (0.8)10 = 0.107374
0 ḷ (10-0)ḷ
- Here m=1, so the binomial formula may be evaluated as:
10 ḷ
P(1) = —————- (0.2)1 (1-0.2)10-1
- ḷ (10-1)ḷ
= 10 x 0.2 x (0.8)9
= 0.268
(c) Here m=5, so the binomial formula may be evaluated as:
10 ḷ
P(5) = —————- (0.2)5 (1-0.2)10-5
5 ḷ (10-5)ḷ
= 0.0264
The 20% abundance of red fragments leads us to expect that finding one red fragment from a sample of 10 would be a likely outcome and conversely, that finding half the sample were red would be much less probable. These answers are consistent with those expectations.
