Math / Data & Statistics

QuestionA chef is trying out a new recipe. From his past experience, it takes him a few trials to get it right. He estimates for the first 15 trials, his probability of getting the recipe right is $35 \%$ in each trial. Since it's new, every trial is considered to be independent from the other. Submit all your answers to three decimal places. (a) The chef would like to try the recipe 10 times, and he wants to add it to the menu if he gets at least 6 right. (i) What is the probability that the new recipe will be added to the menu? 0.021 (ii) What is the probability that the chef gets exactly 6 right? (b) The chef would like to try the recipe 15 times. (i) What is the probability that the chef gets exactly 6 right?

Studdy Solution

STEP 1

What is this asking? What's the chance this chef gets his new recipe right at least 6 times out of 10 tries, and exactly 6 times out of 10 and 15 tries, if he has a 35% chance of success each time? Watch out! Don't mix up "at least" with "exactly"!
Also, make sure to use the right number of trials for each question.

STEP 2

1. Define the probability
2. At least 6 out of 10
3. Exactly 6 out of 10
4. Exactly 6 out of 15

STEP 3

The probability of the chef getting the recipe right in a single trial is $\text{P(Success)} = 35\% = 0.35$ .
Since each trial is independent, the probability stays the same for each trial.

STEP 4

The probability of failure is $\text{P(Failure)} = 1 - \text{P(Success)} = 1 - 0.35 = 0.65$ .

STEP 5

We'll use the binomial probability formula: $P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ , where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success in a single trial.

STEP 6

We want $P(X \ge 6)$ , which means the probability of getting the recipe right 6, 7, 8, 9, or 10 times.
We calculate each probability and add them up!

STEP 7

$P(X=6) = \binom{10}{6} \cdot (0.35)^6 \cdot (0.65)^4 \approx 0.0689$

STEP 8

$P(X=7) = \binom{10}{7} \cdot (0.35)^7 \cdot (0.65)^3 \approx 0.0212$

STEP 9

$P(X=8) = \binom{10}{8} \cdot (0.35)^8 \cdot (0.65)^2 \approx 0.0043$

STEP 10

$P(X=9) = \binom{10}{9} \cdot (0.35)^9 \cdot (0.65)^1 \approx 0.0005$

STEP 11

$P(X=10) = \binom{10}{10} \cdot (0.35)^{10} \cdot (0.65)^0 \approx 0.00003$

STEP 12

$P(X \ge 6) \approx 0.0689 + 0.0212 + 0.0043 + 0.0005 + 0.00003 \approx 0.0949$

STEP 13

We already calculated this in the previous section: $P(X=6) \approx 0.0689$ .

STEP 14

Now, $n = 15$ and $k = 6$ . $P(X=6) = \binom{15}{6} \cdot (0.35)^6 \cdot (0.65)^9 \approx 0.1419$

STEP 15

(a)(i) The probability the recipe is added to the menu is approximately $\textbf{0.095}$ . (a)(ii) The probability the chef gets it right exactly 6 times out of 10 is approximately $\textbf{0.069}$ . (b)(i) The probability the chef gets it right exactly 6 times out of 15 is approximately $\textbf{0.142}$ .

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Studdy Solution

STEP 1

STEP 2

1. Define the probability
2. At least 6 out of 10
3. Exactly 6 out of 10
4. Exactly 6 out of 15

STEP 3

The probability of the chef getting the recipe right in a single trial is $\text{P(Success)} = 35\% = 0.35$ .
Since each trial is independent, the probability stays the same for each trial.

STEP 4

The probability of failure is $\text{P(Failure)} = 1 - \text{P(Success)} = 1 - 0.35 = 0.65$ .

STEP 5

We'll use the binomial probability formula: $P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ , where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success in a single trial.

STEP 6

We want $P(X \ge 6)$ , which means the probability of getting the recipe right 6, 7, 8, 9, or 10 times.
We calculate each probability and add them up!

STEP 7

$P(X=6) = \binom{10}{6} \cdot (0.35)^6 \cdot (0.65)^4 \approx 0.0689$

STEP 8

$P(X=7) = \binom{10}{7} \cdot (0.35)^7 \cdot (0.65)^3 \approx 0.0212$

STEP 9

$P(X=8) = \binom{10}{8} \cdot (0.35)^8 \cdot (0.65)^2 \approx 0.0043$

STEP 10

$P(X=9) = \binom{10}{9} \cdot (0.35)^9 \cdot (0.65)^1 \approx 0.0005$

STEP 11

$P(X=10) = \binom{10}{10} \cdot (0.35)^{10} \cdot (0.65)^0 \approx 0.00003$

STEP 12

$P(X \ge 6) \approx 0.0689 + 0.0212 + 0.0043 + 0.0005 + 0.00003 \approx 0.0949$

STEP 13

We already calculated this in the previous section: $P(X=6) \approx 0.0689$ .

STEP 14

Now, $n = 15$ and $k = 6$ . $P(X=6) = \binom{15}{6} \cdot (0.35)^6 \cdot (0.65)^9 \approx 0.1419$

STEP 15

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Studdy Solution

STEP 1

STEP 2

1. Define the probability2. At least 6 out of 103. Exactly 6 out of 104. Exactly 6 out of 15

STEP 3

The probability of the chef getting the recipe right in a single trial is P(Success)=35%=0.35\text{P(Success)} = 35\% = 0.35P(Success)=35%=0.35.Since each trial is independent, the probability stays the same for each trial.

STEP 4

The probability of failure is P(Failure)=1−P(Success)=1−0.35=0.65\text{P(Failure)} = 1 - \text{P(Success)} = 1 - 0.35 = 0.65P(Failure)=1−P(Success)=1−0.35=0.65.

STEP 5

We'll use the binomial probability formula: P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}P(X=k)=(kn​)⋅pk⋅(1−p)n−k, where nnn is the number of trials, kkk is the number of successes, and ppp is the probability of success in a single trial.

STEP 6

We want P(X≥6)P(X \ge 6)P(X≥6), which means the probability of getting the recipe right 6, 7, 8, 9, or 10 times.We calculate each probability and add them up!

STEP 7

P(X=6)=(106)⋅(0.35)6⋅(0.65)4≈0.0689P(X=6) = \binom{10}{6} \cdot (0.35)^6 \cdot (0.65)^4 \approx 0.0689P(X=6)=(610​)⋅(0.35)6⋅(0.65)4≈0.0689

STEP 8

P(X=7)=(107)⋅(0.35)7⋅(0.65)3≈0.0212P(X=7) = \binom{10}{7} \cdot (0.35)^7 \cdot (0.65)^3 \approx 0.0212P(X=7)=(710​)⋅(0.35)7⋅(0.65)3≈0.0212

STEP 9

P(X=8)=(108)⋅(0.35)8⋅(0.65)2≈0.0043P(X=8) = \binom{10}{8} \cdot (0.35)^8 \cdot (0.65)^2 \approx 0.0043P(X=8)=(810​)⋅(0.35)8⋅(0.65)2≈0.0043

STEP 10

P(X=9)=(109)⋅(0.35)9⋅(0.65)1≈0.0005P(X=9) = \binom{10}{9} \cdot (0.35)^9 \cdot (0.65)^1 \approx 0.0005P(X=9)=(910​)⋅(0.35)9⋅(0.65)1≈0.0005

STEP 11

P(X=10)=(1010)⋅(0.35)10⋅(0.65)0≈0.00003P(X=10) = \binom{10}{10} \cdot (0.35)^{10} \cdot (0.65)^0 \approx 0.00003P(X=10)=(1010​)⋅(0.35)10⋅(0.65)0≈0.00003

STEP 12

P(X≥6)≈0.0689+0.0212+0.0043+0.0005+0.00003≈0.0949P(X \ge 6) \approx 0.0689 + 0.0212 + 0.0043 + 0.0005 + 0.00003 \approx 0.0949P(X≥6)≈0.0689+0.0212+0.0043+0.0005+0.00003≈0.0949

STEP 13

We already calculated this in the previous section: P(X=6)≈0.0689P(X=6) \approx 0.0689P(X=6)≈0.0689.

STEP 14

Now, n=15n = 15n=15 and k=6k = 6k=6. P(X=6)=(156)⋅(0.35)6⋅(0.65)9≈0.1419P(X=6) = \binom{15}{6} \cdot (0.35)^6 \cdot (0.65)^9 \approx 0.1419P(X=6)=(615​)⋅(0.35)6⋅(0.65)9≈0.1419

STEP 15

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1. Define the probability
2. At least 6 out of 10
3. Exactly 6 out of 10
4. Exactly 6 out of 15

The probability of the chef getting the recipe right in a single trial is $\text{P(Success)} = 35\% = 0.35$ .
Since each trial is independent, the probability stays the same for each trial.

The probability of failure is $\text{P(Failure)} = 1 - \text{P(Success)} = 1 - 0.35 = 0.65$ .

We'll use the binomial probability formula: $P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ , where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success in a single trial.

We want $P(X \ge 6)$ , which means the probability of getting the recipe right 6, 7, 8, 9, or 10 times.
We calculate each probability and add them up!

$P(X=6) = \binom{10}{6} \cdot (0.35)^6 \cdot (0.65)^4 \approx 0.0689$

$P(X=7) = \binom{10}{7} \cdot (0.35)^7 \cdot (0.65)^3 \approx 0.0212$

$P(X=8) = \binom{10}{8} \cdot (0.35)^8 \cdot (0.65)^2 \approx 0.0043$

$P(X=9) = \binom{10}{9} \cdot (0.35)^9 \cdot (0.65)^1 \approx 0.0005$

$P(X=10) = \binom{10}{10} \cdot (0.35)^{10} \cdot (0.65)^0 \approx 0.00003$

$P(X \ge 6) \approx 0.0689 + 0.0212 + 0.0043 + 0.0005 + 0.00003 \approx 0.0949$

We already calculated this in the previous section: $P(X=6) \approx 0.0689$ .

Now, $n = 15$ and $k = 6$ . $P(X=6) = \binom{15}{6} \cdot (0.35)^6 \cdot (0.65)^9 \approx 0.1419$