Question

(d) Let p represent the probability of a successful long hit. What values of p will make the long hit better than the short hit in terms of improving the expected value of the score? Explain your reasoning.

Answer 1

Answer:

The probability of successful  long hit is greater than 0.7083

Step-by-step explanation:

Solution

The expected value for long hit is given below

P (successful hit) * 4.2 + P (Unsuccessful hit) *5.4

so,

= P * 4.2 + (1-p) * 5.4

= 5.4 - 1.2p

Now,

If the expected value of long hit is less than  that of the short it

Then,

5.4 - 1.2p <4.55

P> (5.4 - 4.55)/1.2

P> 0.7083

Therefore the probability of successful  long hit is greater than 0.7083

Note:

Due to my research and findings to this example, the complete question for  is attached below, i solved or Question 2(d) only

Answer 2

Probabilities are used to determine the chances of events

The value of p must be greater than 0.71

Start by calculating the expected value of short hit using:

$E(x) = \sum x \times P(x)$

So, we have:

$E(x) = 3 \times 0.15 +4 \times 0.40 + 5 \times 0.25 + 6 \times 0.15 + 7 \times 0.05$

$E(x) = 4.55$

Let the probability of a long hit be p.

Such that the count of a successful hit is 4.2, and unsuccessful hits are 5.4.

So, the expected value is:

$E(x) = \sum x \times P(x)$

$E(p) = 4.2 \times p + (1 -p) \times 5.4$

Expand

$E(p) = 4.2p + 5.4 -5.4p$

Evaluate like terms

$E(p) = 5.4 -1.2p$

This value is less than 4.55; the expected value of short hit.

So, we have:

$5.4 -1.2p < 4.55$

Subtract 5.4 from both sides

$-1.2p < -0.85$

Divide both sides by -1.2

$p > 0.71$

Hence, the value of p must be greater than 0.71 (and less than or equal to 1).

Read more about expected values and probabilities at:

https://brainly.com/question/13934271