10.4 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICSTheoretical Distribution
(c) Probability of 6 heads=
660 60
61 11 0.016
Cpq 22
=× =⎛⎞⎛⎞⎜⎜⎟⎟⎟⎟
⎜⎜⎜⎜⎝⎠⎝⎠⎟⎟⎟⎟
(d) Probability of getting 4 or more heads
= 0.234 + 0.094 + 0.016 = 0.344
Alternatively
Probability of getting at least 4 heads (means we may get 4 heads or 5 heads or 6 heads)
p(x = 4) = P(4) + P(5) + P(6)=
66642 5 60
45611 11 11
CCC 22 22 22
⎛⎞⎛⎞ ⎛⎞⎛⎞ ⎛⎞⎛⎞⎜⎜ ⎜⎜ ⎜⎜⎟⎟ ⎟⎟ ⎟⎟⎟⎟ ⎟⎟ ⎟⎟++
⎜⎜ ⎜⎜ ⎜⎜⎜⎜ ⎜⎜ ⎜⎜⎝⎠⎝⎠ ⎝⎠⎝⎠ ⎝⎠⎝⎠⎟⎟ ⎟⎟ ⎟⎟⎟⎟ ⎟⎟ ⎟⎟
=
666666
456111
CCC 222
⎛⎞ ⎛⎞ ⎛⎞⎜⎜⎜⎟⎟⎟⎟⎟⎟++
⎜⎜⎜⎜⎜⎜⎝⎠ ⎝⎠ ⎝⎠⎟⎟⎟⎟⎟⎟
= ()
6(^1666456)
2 CCC
⎛⎞⎜ ⎟⎟ ++
⎜⎜⎝⎠⎟⎟
= 641 ×++()15 6 1
= 641 ×=22 0.344
Example 2 :
The average percentage of failure in a certain examination is 40. What is the probability that out of a group
of 6 candidates, at least 4 passed in the examination?Solution :
All the trials are independent. The number of pass in the examination may be minimum 4 or 5 or all of them
may pass.
P = 1 – qq = 10040 = 0.4
p = 1 – 0.4 = 0.60
The probability of passing 4 or more candidatesi.e., P(x (^) ≥4) = P(x = 4) + P (x = 5) + P (x = 6)
= P(4) + P(5) + P(6)
=^6 C 4 (0.6)^4 (0.4)^2 +^6 C 5 (0.6)^5 (.4) +^6 C 6 (0.6)^6
= (15 ×0.1296 ×0.16) + (6 ×0.07776 × 0.4) + (0.046656)
= 0.311040 + 0.186624 + 0.046656
= 0.544320