John plays a roulette 5 times; each time betting on red. The probability of red is 1838=919. Let X be the number of wins. Find the expected value (mean) E{X}, the standard deviation, and P(X=2).

Step by step solution to "John plays a roulette 5 times; each time..."

Kye

2021-09-19

John plays a roulette 5 times; each time betting on red. The probability of red is

\frac{18}{38} = \frac{9}{19}

. Let X be the number of wins. Find the expected value (mean) E{X}, the standard deviation, and

P (X = 2)

estenutC

Skilled2021-09-20Added 81 answers

Step 1
Given: John plays a roulette 5 times; each time betting on red. The probability of red is

\frac{18}{38} = \frac{9}{19}

.
Let X be the number of wins.
Number of times plays a roulette:

n = 5

Probability of red:

p = \frac{9}{19} \approx 0.47

Since, all trials are independent and probability of red in each trials is same which is 0.47. Therefore ,

X \sim B (n = 5, p = 0.47)

Binomial probability formula is written as:

P (X = x) = (b e g \in {a r r a y} {c} n ∖ x e n d {a r r a y}) \cdot p^{x} \cdot {(1 - p)}^{n - x}

Step 2
The mean of binomial distribution is :

E (X) = n p

= 5 (0.47)

E (X) = 2.35

The standard deviation of binomial distribution is:

S D (X) = \sqrt{n p (1 - p)}

= \sqrt{5 (0.47) (1 - 0.47)}

S D (X) = 1.1160

Here,

n = 5, p = 0.47 and X = 2

Using binomial probability formula:

P (X = 2) = (b e g \in {a r r a y} {c} 5 ∖ 2 e n d {a r r a y}) \cdot {(0.47)}^{2} \cdot {(1 - 0.47)}^{5 - 2}

= 10 \cdot (0.2209) \cdot (0.1489)

P (X = 2) = 0.3289

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