\[{ 0.267} \]

UpStudy Free Solution:

Given:

- The probability of a marshmallow traveling at least 2 meters (\(p\)) is 0.63.

- The number of trials (\(n\)) is 5.

- We want to find the probability that fewer than 3 marshmallows travel at least 2 meters.

First, we define the random variable \(X\) as the number of marshmallows that travel at least 2 meters. \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = 0.63\).

We need to calculate \(P( X < 3) \), which is the sum of the probabilities of \(X\) being 0, 1, or 2. The probability mass function of a binomial distribution is given by:

\[P( X = k) = \binom { n} { k} p^ k ( 1- p) ^ { n- k} \]

We calculate \(P( X = 0) \), \(P( X = 1) \), and \(P( X = 2) \):

1.For \(X = 0\):

\[P( X = 0) = \binom { 5} { 0} ( 0.63) ^ 0 ( 0.37) ^ 5 = 1 \cdot 1 \cdot ( 0.37) ^ 5 = ( 0.37) ^ 5\]

\[( 0.37) ^ 5 \approx 0.0069\]

2. For \(X = 1\):

\[P( X = 1) = \binom { 5} { 1} ( 0.63) ^ 1 ( 0.37) ^ 4 = 5 \cdot 0.63 \cdot ( 0.37) ^ 4\]

\[5 \cdot 0.63 \cdot ( 0.37) ^ 4 \approx 5 \cdot 0.63 \cdot 0.0187 \approx 0.0590\]

3. For \(X = 2\):

\[P( X = 2) = \binom { 5} { 2} ( 0.63) ^ 2 ( 0.37) ^ 3 = 10 \cdot ( 0.63) ^ 2 \cdot ( 0.37) ^ 3\]

\[10 \cdot ( 0.63) ^ 2 \cdot ( 0.37) ^ 3 \approx 10 \cdot 0.3969 \cdot 0.0507 \approx 0.2010\]

Now, we sum these probabilities to find \(P( X < 3) \):

\[P( X < 3) = P( X = 0) + P( X = 1) + P( X = 2) \]

\[P( X < 3) \approx 0.0069 + 0.0590 + 0.2010 = 0.2669\]

Thus, the probability that fewer than 3 marshmallows will travel at least 2 meters is approximately \(0.267\) when rounded to the nearest thousandth.

Supplemental Knowledge

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It is defined by two parameters:

- \(n\): the number of trials

- \(p\): the probability of success on each trial

The probability mass function (PMF) for a binomial random variable \(X\) representing the number of successes in \(n\) trials is given by:

\[P( X = k) = \binom { n} { k} p^ k ( 1- p) ^ { n- k} \]

where:

- \(\binom { n} { k} = \frac { n! } { k! ( n- k) ! } \) is the binomial coefficient, which counts the number of ways to choose \(k\) successes out of \(n\) trials.

- \(p^ k\): the probability of having exactly \(k\) successes.

- \(( 1- p) ^ { n- k} \): the probability of having exactly \(n- k\) failures.

To find cumulative probabilities, such as \(P( X < 3) \), we sum individual probabilities for all relevant values:

\[P( X < 3) = P( X = 0) + P( X = 1) + P( X = 2) \]

Mastery of probability calculations using distributions like the binomial distribution can be critical in making informed decisions across many fields, from finance and engineering to data analytics and experiment planning. Mastering this tool gives an important competitive edge.At UpStudy, our aim is to assist in understanding complex probability concepts quickly. 

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