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

Question

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

Audrey Allen · Accepted Answer

#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#

Emery Adams · Answer

#:. 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#.

Grace Agate · Answer

undefined The probability $ P(X = 5) $ for a binomial random variable $ X $ with parameters $ p = 0.4 $ and $ n = 10 $ can be calculated using the binomial probability formula:

$$ P(X = k) = \binom{n}{k} \times p^k \times (1 - p)^{n - k} $$

Substitute $ k = 5 $, $ p = 0.4 $, and $ n = 10 $ into the formula:

$$ P(X = 5) = \binom{10}{5} \times (0.4)^5 \times (0.6)^5 $$

Calculate the binomial coefficient $ \binom{10}{5} $ using the formula:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

$$ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 $$

Plug in the values and compute:

$$ P(X = 5) = 252 \times (0.4)^5 \times (0.6)^5 $$

$$ P(X = 5) = 252 \times 0.01024 \times 0.07776 $$

$$ P(X = 5) = 252 \times 0.0007969 $$

$$ P(X = 5) \approx 0.20051 $$

So, $ P(X = 5) $ is approximately $ 0.20051 $ or $ 20.05\% $.

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