Math / Data & Statistics

Questionetting Chrome as default Set as default Module Knowlsdge Chack Question 5
Suppose that $55 \%$ of all babies born in a particular hospital are boys. If 6 babies born in the hospital are randomly selected, what is the probability that fewer than 2 of them are boys?
Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places. (If necessary, consult a list of formulas.) $\square$

Studdy Solution

STEP 1

What is this asking? If 55% of babies born are boys, what's the chance that less than 2 out of 6 randomly chosen newborns are boys? Watch out! "Fewer than 2" means 0 or 1, not 2!
Also, don't forget to round correctly at the end.

STEP 2

1. Define the binomial probability formula
2. Calculate the probability of 0 boys
3. Calculate the probability of 1 boy
4. Calculate the final probability

STEP 3

Hey awesome students!
We're diving into binomial probability!
This formula helps us find the chance of something happening a specific number of times in a set number of trials, like how many boys we see out of 6 babies.

STEP 4

The formula is: $P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{(n-k)}$ Where $P(X=k)$ is the probability of getting exactly $k$ successes (in our case, boys!), $n$ is the total number of trials (6 babies), $k$ is the number of successes we want (0 or 1 boy), and $p$ is the probability of success on a single trial (55% chance of a boy).

STEP 5

Let's plug in the values for 0 boys ( $k=0$ ): $P(X=0) = \binom{6}{0} \cdot (0.55)^0 \cdot (1-0.55)^{(6-0)}$

STEP 6

Remember, anything raised to the power of 0 is 1!
So $(0.55)^0 = 1$ .
Also, $\binom{6}{0} = 1$ because there's only one way to choose 0 boys out of 6.
This simplifies to: $P(X=0) = 1 \cdot 1 \cdot (0.45)^6$

STEP 7

Calculating $(0.45)^6$ gives us approximately 0.0083.
So: $P(X=0) \approx 0.0083$

STEP 8

Now, let's plug in the values for 1 boy ( $k=1$ ): $P(X=1) = \binom{6}{1} \cdot (0.55)^1 \cdot (1-0.55)^{(6-1)}$

STEP 9

We know $\binom{6}{1} = 6$ because there are six ways to choose 1 boy out of 6.
This simplifies to: $P(X=1) = 6 \cdot 0.55 \cdot (0.45)^5$

STEP 10

Calculating $(0.45)^5$ gives us approximately 0.0185.
Now we have: $P(X=1) = 6 \cdot 0.55 \cdot 0.0185$

STEP 11

Multiplying these values gives us approximately 0.0611.
So: $P(X=1) \approx 0.0611$

STEP 12

Since we want the probability of fewer than 2 boys, we add the probabilities of 0 boys and 1 boy: $P(X<2) = P(X=0) + P(X=1)$

STEP 13

Plugging in our calculated values: $P(X<2) \approx 0.0083 + 0.0611$

STEP 14

Adding these probabilities gives us approximately 0.0694.

STEP 15

Rounding to two decimal places, the probability of fewer than 2 boys is approximately 0.07.

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Studdy Solution

STEP 1

What is this asking? If 55% of babies born are boys, what's the chance that less than 2 out of 6 randomly chosen newborns are boys? Watch out! "Fewer than 2" means 0 or 1, not 2!
Also, don't forget to round correctly at the end.

STEP 2

1. Define the binomial probability formula
2. Calculate the probability of 0 boys
3. Calculate the probability of 1 boy
4. Calculate the final probability

STEP 3

Hey awesome students!
We're diving into binomial probability!
This formula helps us find the chance of something happening a specific number of times in a set number of trials, like how many boys we see out of 6 babies.

STEP 4

STEP 5

Let's plug in the values for 0 boys ( $k=0$ ): $P(X=0) = \binom{6}{0} \cdot (0.55)^0 \cdot (1-0.55)^{(6-0)}$

STEP 6

Remember, anything raised to the power of 0 is 1!
So $(0.55)^0 = 1$ .
Also, $\binom{6}{0} = 1$ because there's only one way to choose 0 boys out of 6.
This simplifies to: $P(X=0) = 1 \cdot 1 \cdot (0.45)^6$

STEP 7

Calculating $(0.45)^6$ gives us approximately 0.0083.
So: $P(X=0) \approx 0.0083$

STEP 8

Now, let's plug in the values for 1 boy ( $k=1$ ): $P(X=1) = \binom{6}{1} \cdot (0.55)^1 \cdot (1-0.55)^{(6-1)}$

STEP 9

We know $\binom{6}{1} = 6$ because there are six ways to choose 1 boy out of 6.
This simplifies to: $P(X=1) = 6 \cdot 0.55 \cdot (0.45)^5$

STEP 10

Calculating $(0.45)^5$ gives us approximately 0.0185.
Now we have: $P(X=1) = 6 \cdot 0.55 \cdot 0.0185$

STEP 11

Multiplying these values gives us approximately 0.0611.
So: $P(X=1) \approx 0.0611$

STEP 12

Since we want the probability of fewer than 2 boys, we add the probabilities of 0 boys and 1 boy: $P(X<2) = P(X=0) + P(X=1)$

STEP 13

Plugging in our calculated values: $P(X<2) \approx 0.0083 + 0.0611$

STEP 14

Adding these probabilities gives us approximately 0.0694.

STEP 15

Rounding to two decimal places, the probability of fewer than 2 boys is approximately 0.07.

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Studdy Solution

STEP 1

What is this asking? If 55% of babies born are boys, what's the chance that less than 2 out of 6 randomly chosen newborns are boys? Watch out! "Fewer than 2" means 0 or 1, not 2!Also, don't forget to round correctly at the end.

STEP 2

1. Define the binomial probability formula2. Calculate the probability of 0 boys3. Calculate the probability of 1 boy4. Calculate the final probability

STEP 3

Hey awesome students!We're diving into binomial probability!This formula helps us find the chance of something happening a specific number of times in a set number of trials, like how many boys we see out of 6 babies.

STEP 4

STEP 5

Let's **plug in** the values for 0 boys (k=0k=0k=0): P(X=0)=(60)⋅(0.55)0⋅(1−0.55)(6−0)P(X=0) = \binom{6}{0} \cdot (0.55)^0 \cdot (1-0.55)^{(6-0)}P(X=0)=(06​)⋅(0.55)0⋅(1−0.55)(6−0)

STEP 6

Remember, anything raised to the power of 0 is 1!So (0.55)0=1(0.55)^0 = 1(0.55)0=1.Also, (60)=1\binom{6}{0} = 1(06​)=1 because there's only one way to choose 0 boys out of 6.This simplifies to: P(X=0)=1⋅1⋅(0.45)6P(X=0) = 1 \cdot 1 \cdot (0.45)^6P(X=0)=1⋅1⋅(0.45)6

STEP 7

Calculating (0.45)6(0.45)^6(0.45)6 gives us approximately **0.0083**.So: P(X=0)≈0.0083P(X=0) \approx 0.0083P(X=0)≈0.0083

STEP 8

Now, let's **plug in** the values for 1 boy (k=1k=1k=1): P(X=1)=(61)⋅(0.55)1⋅(1−0.55)(6−1)P(X=1) = \binom{6}{1} \cdot (0.55)^1 \cdot (1-0.55)^{(6-1)}P(X=1)=(16​)⋅(0.55)1⋅(1−0.55)(6−1)

STEP 9

We know (61)=6\binom{6}{1} = 6(16​)=6 because there are six ways to choose 1 boy out of 6.This simplifies to: P(X=1)=6⋅0.55⋅(0.45)5P(X=1) = 6 \cdot 0.55 \cdot (0.45)^5P(X=1)=6⋅0.55⋅(0.45)5

STEP 10

Calculating (0.45)5(0.45)^5(0.45)5 gives us approximately **0.0185**.Now we have: P(X=1)=6⋅0.55⋅0.0185P(X=1) = 6 \cdot 0.55 \cdot 0.0185P(X=1)=6⋅0.55⋅0.0185

STEP 11

Multiplying these values gives us approximately **0.0611**.So: P(X=1)≈0.0611P(X=1) \approx 0.0611P(X=1)≈0.0611

STEP 12

Since we want the probability of *fewer* than 2 boys, we **add** the probabilities of 0 boys and 1 boy: P(X<2)=P(X=0)+P(X=1)P(X<2) = P(X=0) + P(X=1)P(X<2)=P(X=0)+P(X=1)

STEP 13

Plugging in our calculated values: P(X<2)≈0.0083+0.0611P(X<2) \approx 0.0083 + 0.0611P(X<2)≈0.0083+0.0611

STEP 14

Adding these probabilities gives us approximately **0.0694**.

STEP 15

Rounding to two decimal places, the probability of fewer than 2 boys is approximately **0.07**.

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? If 55% of babies born are boys, what's the chance that less than 2 out of 6 randomly chosen newborns are boys? Watch out! "Fewer than 2" means 0 or 1, not 2!
Also, don't forget to round correctly at the end.

1. Define the binomial probability formula
2. Calculate the probability of 0 boys
3. Calculate the probability of 1 boy
4. Calculate the final probability

Hey awesome students!
We're diving into binomial probability!
This formula helps us find the chance of something happening a specific number of times in a set number of trials, like how many boys we see out of 6 babies.

Let's plug in the values for 0 boys ( $k=0$ ): $P(X=0) = \binom{6}{0} \cdot (0.55)^0 \cdot (1-0.55)^{(6-0)}$

Remember, anything raised to the power of 0 is 1!
So $(0.55)^0 = 1$ .
Also, $\binom{6}{0} = 1$ because there's only one way to choose 0 boys out of 6.
This simplifies to: $P(X=0) = 1 \cdot 1 \cdot (0.45)^6$

Calculating $(0.45)^6$ gives us approximately 0.0083.
So: $P(X=0) \approx 0.0083$

Now, let's plug in the values for 1 boy ( $k=1$ ): $P(X=1) = \binom{6}{1} \cdot (0.55)^1 \cdot (1-0.55)^{(6-1)}$

We know $\binom{6}{1} = 6$ because there are six ways to choose 1 boy out of 6.
This simplifies to: $P(X=1) = 6 \cdot 0.55 \cdot (0.45)^5$

Calculating $(0.45)^5$ gives us approximately 0.0185.
Now we have: $P(X=1) = 6 \cdot 0.55 \cdot 0.0185$

Multiplying these values gives us approximately 0.0611.
So: $P(X=1) \approx 0.0611$

Since we want the probability of fewer than 2 boys, we add the probabilities of 0 boys and 1 boy: $P(X<2) = P(X=0) + P(X=1)$

Plugging in our calculated values: $P(X<2) \approx 0.0083 + 0.0611$

Adding these probabilities gives us approximately 0.0694.

Rounding to two decimal places, the probability of fewer than 2 boys is approximately 0.07.