Top half:
D (x-2, y+3) E (x-3, y(+\-) 0) F (x+1, y-2)
P (x+1, y+2) Q (x+3, y+4) R (x+5, y+2)
The answer is 8/27 just so you know
Answer:
(a) The probability of more than one death in a corps in a year is 0.1252.
(b) The probability of no deaths in a corps over 7 years is 0.0130.
Step-by-step explanation:
Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.
The random variable
.
The probability function of a Poisson distribution is:

(a)
Compute the probability of more than one death in a corps in a year as follows:
P (X > 1) = 1 - P (X ≤ 1)
= 1 - P (X = 0) - P (X = 1)

Thus, the probability of more than one death in a corps in a year is 0.1252.
(b)
The average deaths over 7 year period is: 
Compute the probability of no deaths in a corps over 7 years as follows:

Thus, the probability of no deaths in a corps over 7 years is 0.0130.
The available balance online may not reflect all debit transactions or all checks that were written. Some recent debit transactions and check written may not have been posted to his account yet. The available balance would need to subtract any outstanding debit transactions and any written checks that have not cleared his account. This would yield the correct available balance.
Answer:
{0.16807, 0.36015, 0.3087, 0.1323, 0.02835, 0.00243}
Step-by-step explanation:
The expansion of (p+q)^n for n = 5 is ...
(p+q)^5 = p^5 +5·p^4·q +10·p^3·q^2 +10·p^2·q^3 +5·p·q^4 +q^5
When the probability p=0.3 and q = 1-p = 0.7 the terms of this series correspond to the probabilities of 5, 4, 3, 2, 1, and 0 favorable outcomes out of 5 trials.
For example, p^5 = 0.3^5 = 0.00243 is the probability of 5 favorable outcomes in 5 trials where the probability of each favorable outcome is 0.3.
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The attachment shows the calculation of these numbers using a graphing calculator. It lists them in reverse order of the expansion of (p+q)^5 shown above, so that they are the probabilities of 0–5 favorable outcomes in the order 0–5.