Question

this friday, tyler has a french vocabulary test as well as a spanish vocabulary test. to prepare, he made a study card for each word, 40% of which are french. every time he picks a card, he sticks it back in the deck and shuffles again. if tyler picks a study card from the deck 5 times during his first study session, what is the probability that at most 1 card has a french word? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success on a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Here, \(n = 5\), \(p=0.4\) (probability of picking a French - word card), and we want \(P(X\leq1)=P(X = 0)+P(X = 1)\)

Step2: Calculate \(P(X = 0)\)

For \(k = 0\):
\(C(5,0)=\frac{5!}{0!(5 - 0)!}=1\)
\(P(X = 0)=C(5,0)\times(0.4)^{0}\times(1 - 0.4)^{5-0}\)
\(P(X = 0)=1\times1\times(0.6)^{5}\)
\(P(X = 0)=0.6^{5}=0.07776\)

Step3: Calculate \(P(X = 1)\)

For \(k = 1\):
\(C(5,1)=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=5\)
\(P(X = 1)=C(5,1)\times(0.4)^{1}\times(0.6)^{4}\)
\(P(X = 1)=5\times0.4\times0.6^{4}\)
\(0.6^{4}=0.1296\), so \(P(X = 1)=5\times0.4\times0.1296 = 0.2592\)

Step4: Calculate \(P(X\leq1)\)

\(P(X\leq1)=P(X = 0)+P(X = 1)\)
\(P(X\leq1)=0.07776+0.2592=0.33696\approx0.337\)

Answer:

\(0.337\)