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Taya2010 [7]
3 years ago
14

Flip a fair coin 5 times. Let random variable X = "number of heads."What is P(X = 1)?

Mathematics
1 answer:
kaheart [24]3 years ago
3 0

Answer:

The probability is 1/32

Step-by-step explanation:

What the question asks here is that in a throw of a coin 5 times, what is the probability that we have the head being revealed once in the 5 throws

for a fair coin, we can only get a head or a tail

So the probability of getting a head equals that of getting a tail

Since total probability is 1; then the probability of each will be 1/2

So, we can now have the probability as follows;

1 head (1/2) and 4 tails (1/2 * 1/2 * 1/2 * 1/2)

1/2 * 1/2 * 1/2 * 1/2 * 1/2 = (1/2)^5 = 1/32

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mart [117]

Answer:

26200 is 655% of 400

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4 years ago
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A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0
3 years ago
A^2+b^2=????????????????????????
zepelin [54]

What you have written here is an equation.

You can define as

a2+b2=y  

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Anettt [7]

Answer:

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Step-by-step explanation:

Given function is (2x+4)^{2}-(2x^{2} +6)

On simplifying,

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