Show that in the space R' the vectors r = (1,1,0), y = (0,1,2) and =
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Answer:
Probability of fewer than 4 = 0.0339
Explanation:
This is a question on binomial distribution. It is a distribution which has two possible outcome; a success or a failure
Probability of success (p) = 49%
Probability of failure (q) = 1 - 49% = 1-0.49
Probability of failure (q) = 0.51 = 51%
n = 14
P(X = x) = (n!)/[(n-x)!x!] × p^x × q^(n-x)
Probability of fewer than 4 = Pr(X <4)
Pr(X <4) = Pr(X = 0) + Pr(X = 1) +Pr(X = 2) +Pr(X = 3)
Find attached the workings
From the workings, Pr(X <4) = 0.0339295
Probability of fewer than 4 = 0.0339 (approximately)
