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20% of the patron's order the chef's special. The probability that 2 out of the next ten customers will order the chef's special is

Sagot :

MrRoyal

Answer:

[tex]P(x =2) = 0.3020[/tex]

Step-by-step explanation:

Given

[tex]p =20\% = 0.20[/tex]

[tex]n = 10[/tex]

Required

[tex]P(x = 2)[/tex]

This question is an illustration of binomial distribution where:

[tex]P(X = x) = ^nC_x * p^x * (1 - p)^{n-x[/tex]

So, we have:

[tex]P(x =2) = ^{10}C_2 * 0.20^2 * (1 - 0.20)^{10-2}[/tex]

[tex]P(x =2) = ^{10}C_2 * 0.20^2 * 0.80^8[/tex]

This gives

[tex]P(x =2) = \frac{10!}{(10 - 2)!2!} * 0.20^2 * 0.80^8[/tex]

[tex]P(x =2) = \frac{10!}{8!2!} * 0.20^2 * 0.80^8[/tex]

Expand

[tex]P(x =2) = \frac{10*9*8!}{8!2*1} * 0.20^2 * 0.80^8[/tex]

[tex]P(x =2) = \frac{10*9}{2} * 0.20^2 * 0.80^8[/tex]

[tex]P(x =2) = 45 * 0.20^2 * 0.80^8[/tex]

[tex]P(x =2) = 0.3020[/tex]

The probability that 2 out of the next ten customers will order the chef special is 0.3020. and this can be determined by using the binomial distribution.

Given :

  • 20% of the patron's order the chef's special.
  • Sample size, n = 10

To determine the probability formula of the binomial distribution is used, that is:

[tex]\rm P(x = r) = \; ^nC_r \times p^r \times (1 - p)^{n-r}[/tex]

Now, at n = 10 and r = 2, the probability is given by:

[tex]\rm P(x = 2) = \; ^{10}C_2 \times (0.2)^2 \times (1 - 0.2)^{10-2}[/tex]

[tex]\rm P(x = 2) = \; ^{10}C_2 \times (0.2)^2 \times (0.8)^{10-2}[/tex]

[tex]\rm P(x = 2) = \; \dfrac{10!}{(10-2)!\times 2!} \times (0.2)^2 \times (0.8)^{8}[/tex]

[tex]\rm P(x = 2) = \; 45 \times (0.2)^2 \times (0.8)^{8}[/tex]

P(x  = 2) = 0.3020

The probability that 2 out of the next ten customers will order the chef special is 0.3020.

For more information, refer to the link given below:

https://brainly.com/question/1957976

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