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3 years ago

The probability that a randomly selected 5-year-old male horned beetle will live to be 6 years old is 0.25624.

Mathematics

2 answers:

dalvyx [7]3 years ago

8 0

Answer:

(a) The probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.06566.

(b) The probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.000283.

Step-by-step explanation:

It is given that the probability that a randomly selected 5-year-old male horned beetle will live to be 6 years old is 0.25624.

(a)

We need to find the probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old.

$P=0.25624\times 0.25624=(0.25624)^2=0.0656589376\approx 0.06566$

Therefore the probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.06566.

(b)

We need to find the probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old.

$P=0.25624\times 0.25624\times 0.25624\times 0.25624\times 0.25624\times 0.25624=(0.25624)^6=0.000283061988948\approx 0.000283$

Therefore the probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.000283.

Vadim26 [7]3 years ago

7 0

Answer: (a) 0.0656

(b) 0.00028

Step-by-step explanation:

Since we have given that

Probability that a randomly selected 5-year old male horned beetle will live to be 6 years old = 0.25624

If there are

Number of male horned beetles = 2

So, probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old is given by

$(0.25624)^2\\\\=0.0656$

If number of male horned beetles = 6

So, probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old is given by

$(0.25624)^6\\\\=0.00028\\$

Hence, (a) 0.0656

(b) 0.00028

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The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

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

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

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

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

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

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

<em>The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.</em>

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3 years ago

The probability that a randomly selected​ 5-year-old male horned beetle will live to be 6 years old is 0.25624.​

The probability that a randomly selected 5-year-old male horned beetle will live to be 6 years old is 0.25624.