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2 years ago

9

at a pizza place there are 4 diffrent sizes of pizza(small medium large and extra large. and 8 diffrent toppings (chicken, pepor

oni mushrooms peppers tomatos onions puneapple ham) whats the probability of ordering a medium pizza with peporoni

1 answer:

Vilka [71]2 years ago

6 0

Answer:

1/32(I'm not sure if this is it)

Step-by-step explanation:

The total possible outcomes have 32 different outcome and there's only one that is medium with peporoni

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Oliga [24]

the answer is -3/2 and -11/2

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2 years ago

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Mama L [17]

Answer:

$(a)\ P(x = 0) = 0.2725$

$(b)\ P(x \ge 1) =0.7275$

$(c)\ P(x \le 2) = 0.8948$

Step-by-step explanation:

Given

$n = 8$ --- 8 friends

$p = 15\%$ --- proportion that one-time fling

This question is an illustration of binomial probability, and it is represented as:

$P(X = x) = ^nC_x* p^x * (1 - p)^{n-x}$

Solving (a): P(x = 0) --- None has done one time fling

$P(x = 0) = ^8C_0* (15\%)^0 * (1 - 15\%)^{8-0}$

$P(x = 0) = 1* 1 * (1 - 0.15)^{8}$

$P(x = 0) = 0.85^8$

$P(x = 0) = 0.2725$

Solving (b): $P(x \ge 1)$

To do this, we make use of compliment rule:

$P(x = 0) + P(x \ge 1) =1$

Rewrite as:

$P(x \ge 1) =1 - P(x = 0)$

$P(x \ge 1) =1 - 0.2725$

$P(x \ge 1) =0.7275$

Solving (c): $P(x\le 2)$ --- Not more than 2 has one time fling

This is calculated as:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

We have:

$P(x = 0) = 0.2725$

$P(x = 1) = ^8C_1* (15\%)^1 * (1 - 15\%)^{8-1}$

$P(x = 1) = 8* (0.15) * (1 - 0.15)^7$

$P(x = 1) = 0.3847$

$P(x = 2) = ^8C_2* (15\%)^2 * (1 - 15\%)^{8-2}$

$P(x = 2) = 28* (0.15)^2 * (1-0.15)^6$

$P(x = 2) = 0.2376$

So:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

$P(x \le 2) = 0.2725 + 0.3847 + 0.2376$

$P(x \le 2) = 0.8948$

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