We model this situation using a binomial distribution with parameters: number of trials \(n = 10\) and probability of success \(p = 0.35\).

The probability that Janice buys at most 7 items is given by:

\(P(X \leq 7) = 1 - P(X = 8) - P(X = 9) - P(X = 10)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 8) = \binom{10}{8} (0.35)^8 (0.65)^2\)

\(P(X = 9) = \binom{10}{9} (0.35)^9 (0.65)^1\)

\(P(X = 10) = \binom{10}{10} (0.35)^{10} (0.65)^0\)

Substituting these into the equation:

\(P(X \leq 7) = 1 - (0.004281 + 0.0005123 + 0.00002759)\)

\(P(X \leq 7) = 1 - 0.00482089\)

\(P(X \leq 7) = 0.995\)

Let the random variable \(X\) represent the number of children who own a bicycle. \(X\) follows a binomial distribution with parameters \(n = 15\) and \(p = 0.65\).

We need to find \(P(X > 12) = P(X = 13) + P(X = 14) + P(X = 15)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 13) = \binom{15}{13} (0.65)^{13} (0.35)^2\)

\(P(X = 14) = \binom{15}{14} (0.65)^{14} (0.35)^1\)

\(P(X = 15) = \binom{15}{15} (0.65)^{15} (0.35)^0\)

Summing these probabilities gives:

\(P(X > 12) = 0.0617\)

Let the probability that a customer shops online be \(p = 0.35\). The number of customers chosen is \(n = 6\). We need to find the probability that more than three customers shop online, i.e., \(P(X > 3)\) where \(X\) is a binomial random variable with parameters \(n = 6\) and \(p = 0.35\).

We calculate \(P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:

\(P(X = 4) = \binom{6}{4} (0.35)^4 (0.65)^2\)

\(P(X = 5) = \binom{6}{5} (0.35)^5 (0.65)^1\)

\(P(X = 6) = \binom{6}{6} (0.35)^6 (0.65)^0\)

Calculating each term:

\(P(X = 4) = 15 \times 0.35^4 \times 0.65^2\)

\(P(X = 5) = 6 \times 0.35^5 \times 0.65\)

\(P(X = 6) = 1 \times 0.35^6\)

Summing these probabilities gives \(P(X > 3) = 0.117\).

Let the random variable \(X\) represent the number of days Jake completes the puzzle in 5 days. \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = 0.75\).

We need to find \(P(X \geq 3)\), which is \(P(X = 3) + P(X = 4) + P(X = 5)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\):

\(P(X = 3) = \binom{5}{3} (0.75)^3 (0.25)^2 = 10 \times 0.421875 \times 0.0625 = 0.26367\)

\(P(X = 4) = \binom{5}{4} (0.75)^4 (0.25)^1 = 5 \times 0.31640625 \times 0.25 = 0.39551\)

\(P(X = 5) = \binom{5}{5} (0.75)^5 (0.25)^0 = 1 \times 0.2373046875 \times 1 = 0.23730\)

Adding these probabilities gives:

\(P(X \geq 3) = 0.26367 + 0.39551 + 0.23730 = 0.896\)

The problem involves a binomial distribution where the probability of passing, \(p\), is 0.8, and the number of trials, \(n\), is 9. We need to find the probability that at most 6 students pass, which is \(P(X \leq 6)\).

This can be calculated as:

\(P(X \leq 6) = 1 - (P(7) + P(8) + P(9))\)

Using the binomial probability formula:

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

Calculate \(P(7)\), \(P(8)\), and \(P(9)\):

\(P(7) = \binom{9}{7} (0.8)^7 (0.2)^2\)

\(P(8) = \binom{9}{8} (0.8)^8 (0.2)^1\)

\(P(9) = \binom{9}{9} (0.8)^9 (0.2)^0\)

Substitute these into the equation:

\(P(X \leq 6) = 1 - (0.3019899 + 0.3019899 + 0.1342177)\)

\(P(X \leq 6) = 1 - 0.7381975 = 0.2618025\)

Rounding to three decimal places, the probability is \(0.262\).