PROBABILITY OF A SEQUENCE OF EVENTS
A box contains 12 bolts. Of these, 8 have square heads and 4 have hexagonal heads. Sev-
en bolts will be removed from the box, individually and at random. What is the probabili-
ty that the second and third bolts drawn will have square heads and the sixth bolt will
have a hexagonal head?
Calculation Procedure:- Compute the total number of ways in which the bolts
can be drawn
The sequence in which the bolts are drawn represents a permutation of 12 bolts taken 7 at
a time, and each bolt is unique. The total number of permutations = P 12 J — 121/5!. - Compute the number of ways in which the bolts can be drawn
in the manner specified
If the bolts are drawn in the manner specified, the second and third positions in the per-
mutation are occupied by square-head bolts and the sixth position is occupied by a hexag-
onal-head bolt. Construct such a permutation, in these steps: Place a square-head bolt in
the second position; the number of bolts available is 8. Now place a square-head bolt in
the third position; the number of bolts available is 7. Now place a hexagonal-head bolt in
the sixth position; the number of bolts available is 4. Finally, fill the four remaining posi-
tions in any manner whatever; the number of bolts available is 9.
The second position can be filled in 8 ways, the third position in 7 ways, the sixth po-
sition in 4 ways, and the remaining positions in P 94 ways. By the multiplication law, the
number of acceptable permutations is 8 x 7 * 4 x P 94 = 224(9!/5!). - Compute the probability of drawing the bolts
in the manner specified
Since all permutations have an equal likelihood of becoming the true permutation, the
probability equals the ratio of the number of acceptable permutations to the total number
of permutations. Thus, probability = 224(9!/5!)/(12!/5!) = 224(9!)/!2! = 224(12 x U x
- = 224/1320-0.1697.
- Compute the probability by an alternative approach
As the preceding calculations show, the exact positions specified (second, third, and
sixth) do not affect the result. For simplicity, assume that the first and second bolts are to
be square-headed and the third bolt hexagonal-headed. The probabilities are: first bolt
square-headed, 8/12; second bolt square-headed, 7/11; third bolt hexagonal-headed, 4/10.
The probability that all three events will occur is the product of their respective probabili-
ties. Thus, the probability that bolts will be drawn in the manner specified =
(8/12)(7/l 1)(4/10) = 224/1320 = 0.1697. Note also that the precise number of bolts drawn
from the box (7) does not affect the result.
PROBABILITY ASSOCIATED
WITH A SERIES OF TRIALS
During its manufacture, a product passes through five departments, A, B, C, D, and E.
The probability that the product will be delayed in a department is: A, 0.06; B, 0.15; C,
0.03; D. 0.07; E, 0.13. These values are independent of one another in the sense that the