Expand the equation (p+q)^5

1 answer:

7 0

(p + q)⁵
(p + q)(p + q)(p + q)(p + q)(p + q)
{[p(p + q) + q(p + q)][p(p + q) + q(p + q)](p + q)}
{[p(p) + p(q) + q(p) + q(q)][p(p) + p(q) + q(p) + q(q)](p + q)}
(p² + pq + pq + q²)(p² + pq + pq + q²)(p + q)
(p² + 2pq + q²)(p² + 2pq + q²)(p + q)
{[p²(p² + 2pq + q²) + 2pq(p² + 2pq + q²) + q²(p² + 2pq + q²)](p + q)}
{[p²(p²) + p²(2pq) + p²(q²) + 2pq(p²) + 2pq(2pq) + 2pq(q²) + q²(p²) + q²(2pq) + q²(q²)](p + q)}
(p⁴ + 2p³q + p²q² + 2p³q + 4p²q² + 2pq³ + p²q² + 2pq³ + q⁴)(p + q)
(p⁴ + 2p³q + 2p³q + p²q² + 4p²q² + p²q² + 2pq³ + 2pq³ + q⁴)(p + q)
(p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴)(p + q)
p⁴(p + q) + 4p³q(p + q) + 6p²q²(p + q) + 4pq³(p + q) + q⁴(p + q)
p⁴(p)+ p⁴(q) + 4p³q(p) + 4p³q(q) + 6p²q²(p) + 6p²q²(q) + 4pq³(p) + 4pq³(q) + q⁴(p) + q⁴(q)
p⁵ + p⁴q + 4p⁴q + 4p³q² + 6p³q² + 6p²q³ + 4p²q³ + 4pq⁴ + pq⁴ + q⁵
p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵

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70% of the students applying to a university are accepted. Assume the requirements for a binomial experiment are satisfied for 1

marshall27 [118]

Answer:

a) 0.3826

b) 0.9894

Step-by-step explanation:

We are given the following information:

We treat students accepted at university as a success.

P(students accepted at university) = 70% = 0.70

Then the number of students accepted at university follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

a) P( 8 or more will be accepted)

$P(x \geq 8) = P(x = 8) + P(x = 9) + P(X=10)\\\\= \binom{10}{8}(0.7)^8(1-0.7)^2 + \binom{10}{9}(0.7)^9(1-0.7)^1+ \binom{10}{10}(0.7)^{10}(1-0.7)^0\\\\= 0.2334 + 0.1210 + 0.0282\\= 0.3826$

b) P(4 or more will be accepted)

$P(x \geq 4) =1 - P(x = 0) - P(x = 1) - P(X=2)-P(x=3)\\\\=1 - ( \binom{10}{0}(0.7)^0(1-0.7)^{10} + \binom{10}{1}(0.7)^1(1-0.7)^9+ \binom{10}{2}(0.7)^{2}(1-0.7)^8+ \binom{10}{3}(0.7)^{3}(1-0.7)^7 )\\\\= 1 - 0.0106\\= 0.9894$