Let #X# be a binomial random variable with #p=0.4# and #n=10#. What is # P(X = 5) #?

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Answer 1 · 2017-02-07T07:46:54

#0.2007#

#P(X=r) = ^n C_r(p)^r(q)^(n-r)#

#p=0.4#, then #q=1-0.4=0.6#

#P(X=5) = ^10C_5(0.4)^5(0.6)^5#

#P(X=5) = (10!)/(5!*5!)(0.4)^5(0.6)^5#

#P(X=5) = 252*0.01024*0.07776=0.2007#

Answer 2 · 2017-02-07T07:56:28

#:. P(X=5)=0.2006581248#.

Let #X# be the Binomial Random Variable with parameters #n, p.#

Then, #P(X=x)=""_nC_xp^xq^(n-x), x=0,1,2,3,..........,n; q=1-p.#

In our Example, #x=5, n=10, and, p=0.4#

#:. P(X=5)=""_10C_5(0.4)^5(1-0.4)^(10-5)#

#={(10)(9)(8)(7)(6)}/{(1)(2)(3)(4)(5)}(0.4)^5(0.6)^5#

#=(252)(0.24)^5#

#=(252)(0.0007962624)#

#:. P(X=5)=0.2006581248#.