Example Question - probability
Here are examples of questions we've helped users solve.
Probability of Family Composition
<p>Sea \( p \) la probabilidad de tener un hijo hombre y \( q \) la probabilidad de tener una hija, donde \( p + q = 1 \).</p> <p>Supongamos que \( p = 0.5 \) y \( q = 0.5 \).</p> <p>La probabilidad de no tener hijos hombres en una familia puede ser calculada como \( P(\text{sin hijos hombres}) = q^n \), donde \( n \) es el número total de hijos.</p> <p>Para \( n = 1 \), \( P(\text{no hay hombres}) = 0.5^1 = 0.5 \). Para \( n = 2 \), \( P(\text{no hay hombres}) = 0.5^2 = 0.25 \). Así sucesivamente.</p> <p>Esto depende del número total de hijos que se consideren.</p>
Poisson Distribution Probability Problem
<p>La probabilidad de que una estación de esquí abra antes de diciembre es del 5%, lo que se puede expresar como $\lambda = 0.05$.</p> <p>Utilizando la distribución de Poisson, la probabilidad de que al menos una estación abra antes de diciembre se calcula como:</p> <p>$P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{e^{-\lambda} \lambda^k}{k!}$, donde $k = 0$.</p> <p>Por lo tanto, al sustituir $\lambda$:</p> <p>$P(X = 0) = e^{-0.05 \cdot 100} \frac{(0.05 \cdot 100)^0}{0!} = e^{-5}.$</p> <p>Finalmente, calculamos:</p> <p>$P(X \geq 1) = 1 - e^{-5}$.</p>
Differentiation of xcosx and Probability of Drawing Specific Cards from a Deck
<p>The image displays two separate questions. I will provide the solutions for both.</p> <p>For the differentiation of \( x\cos{x} \) with respect to \( x \) using the first principle:</p> <p>We have \( f(x) = x\cos{x} \), we need to find \( f'(x) \) using the first principle:</p> <p>\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]</p> <p>\[ = \lim_{{h \to 0}} \frac{(x+h)\cos(x+h) - x\cos{x}}{h} \]</p> <p>We then expand and arrange the expression and apply the limit. However, without the options or further context for finding the value of \( k \), this part of the question is incomplete.</p> <p>For the probability question, assuming we are looking to find the probability of drawing 1 diamond and 3 spades:</p> <p>The total number of ways to draw 4 cards from a 52 card deck is \( C(52, 4) \).</p> <p>The number of ways to draw 1 diamond from the 13 available diamonds is \( C(13, 1) \).</p> <p>The number of ways to draw 3 spades from the 13 available spades is \( C(13, 3) \).</p> <p>The probability \( P \) of the event is:</p> <p>\[ P = \frac{C(13, 1) \cdot C(13, 3)}{C(52, 4)} \]</p> <p>\[ P = \frac{13 \cdot \frac{13!}{3!(13-3)!}}{\frac{52!}{4!(52-4)!}} \]</p> <p>We can then simplify the factorials to get the probability.</p>
Probability of Successfully Opening a Lock with Multiple Keys
<p>The first part of the problem states that there is a \(\frac{1}{3}\) probability of opening the lock when Asri chooses a key randomly. Therefore, the probability that Asri fails to open the lock is:</p> <p>\[1 - \frac{1}{3} = \frac{2}{3}\]</p> <p>(a) The probability that Asri fails to open the lock on her first attempt:</p> <p>\[\frac{2}{3}\]</p> <p>(b) Asri has two new keys added to her keychain that cannot open the lock. Let \(n\) be the total number of keys after adding two ineffective keys, then:</p> <p>\[n = 3 + 2 = 5\]</p> <p>The probability that she chooses a key that can open the lock is still \(\frac{1}{3}\), since the number of working keys has not changed. The total number of keys is 5 now, so the probability she chooses a working key is:</p> <p>\[\frac{1}{5} \times \frac{1}{3}\]</p> <p>And the probability she fails to open the lock with the new keychain is:</p> <p>\[1 - \frac{1}{5} \times \frac{1}{3}\]</p> <p>\[\Rightarrow 1 - \frac{1}{15}\]</p> <p>\[\Rightarrow \frac{15}{15} - \frac{1}{15}\]</p> <p>\[\Rightarrow \frac{14}{15}\]</p> <p>The new probability that the lock is successfully opened is:</p> <p>\[\frac{1}{15}\]</p>
Probability of Failing and Succeeding in Opening a Lock with Multiple Keys
<p>(a) The probability that Asri fails to open the lock:</p> <p>P(\text{failure}) = 1 - P(\text{success})</p> <p>P(\text{failure}) = 1 - \frac{1}{3}</p> <p>P(\text{failure}) = \frac{2}{3}</p> <p>(b) The probability that Asri fails to open the lock with the new key added:</p> <p>Since the two new keys cannot open the lock, the total number of keys is now 5.</p> <p>P(\text{new failure}) = \frac{\text{Number of keys that do not work}}{\text{Total number of keys}}</p> <p>P(\text{new failure}) = \frac{4}{5}</p> <p>(c) The probability that the lock is successfully opened with the new key added:</p> <p>P(\text{new success}) = 1 - P(\text{new failure})</p> <p>P(\text{new success}) = 1 - \frac{4}{5}</p> <p>P(\text{new success}) = \frac{1}{5}</p>
Understanding the Null Hypothesis in Probability of Gender Distribution
<p>The null hypothesis states that each child has an equal chance of being a girl, which is a probability of 0.5. Since there are four children, the probability that all children are girls or all are boys (the two extreme cases) is $0.5^4 = 0.0625$ for each case. Together, the probability for either all girls or all boys is $0.0625 + 0.0625 = 0.125$, hence $H_0: p = 0.125$.</p> <p>The test statistic for a proportion is $Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size.</p> <p>Here, $\hat{p} = \frac{21}{120}$, $p_0 = 0.125$, and $n = 120$. So:</p> <p>$Z = \frac{\frac{21}{120} - 0.125}{\sqrt{\frac{0.125(1-0.125)}{120}}}$</p> <p>Compute the Z-value and then compare it to the critical value for a one-tailed test at the 10% significance level, which is $Z = 1.28$ for $Z > 0$ (looking at a standard Z-table and finding the value corresponding to 0.9 since it is one-tailed).</p> <p>If the calculated $Z$-value is greater than the critical value, we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.</p>
Assessment of an Amateur Swimmer's Performance and Statistical Analysis
<p>(a)</p> <p>To find the probability that Zhou wins a randomly chosen race, we use the area of the overlapping section of two normal distributions.</p> <p>Let \( Z_Z \) be the standard normal variable for Zhou's time,</p> \[ Z_Z = \frac{X - 80}{2} \] <p>Let \( Z_T \) be the standard normal variable for Tan's time,</p> \[ Z_T = \frac{X - 79}{3} \] <p>We need P(\( Z_Z < Z_T \)) which is equivalent to P(\( X_Z < X_T \)).</p> <p>Let \( D = X_T - X_Z \), where \( D \) follows N(1, \( 2^2 + 3^2 \)) since \( Var(X_T - X_Z) = Var(X_T) + Var(X_Z) \) as they are independent.</p> <p>Hence, \( D \) ~ N(1, 13).</p> <p>We can standardize \( D \) to get \( Z_D \) ~ N(0, 1) and find P(\( Z_D > 0 \)) to find the probability that Tan wins:</p> \[ Z_D = \frac{D - 1}{\sqrt{13}} \] \[ P(Z_D > 0) = P\left(\frac{D - 1}{\sqrt{13}} > \frac{0 - 1}{\sqrt{13}}\right) \] \[ P(Z_D > 0) = P(Z_D > -1/\sqrt{13}) \] <p>Using standard normal tables, find P(\( Z_D > -0.277 \)).</p> <p>The probability that Zhou wins is the complement of this probability:</p> \[ P(Zhou\ wins) = 1 - P(Tan\ wins) \] \[ P(Zhou\ wins) = 1 - P(Z_D > -0.277) \] <p>(b)</p> <p>Population mean estimate (unbiased) for Zhou's times (µ̂):</p> \[ \mû = \bar{X} = \frac{\sum{X}}{n} = \frac{2376.3}{30} \] <p>Population variance estimate for Zhou's times (σ̂²):</p> \[ \sigmâ^2 = \frac{\sum{X^2} - \frac{(\sum{X})^2}{n}}{n-1} \] \[ \sigmâ^2 = \frac{188653.7 - \frac{(2376.3)^2}{30}}{30 - 1} \] <p>(c)</p> <p>Null Hypothesis \( H_0 \): Zhou's mean time has not reduced, \( \mu = \mu_0 \), where \( \mu_0 \) is the mean time before the exercise regime.</p> <p>Alternative Hypothesis \( H_1 \): Zhou's mean time has reduced, \( \mu < \mu_0 \).</p> <p>The test statistic for a left-tailed t-test, since \( n \) is small and population variance is unknown, is:</p> \[ t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \] <p>Use t-tables to find critical value for α = 0.05 and \( n-1 \) degrees of freedom. If \( t < t_{critical} \), reject \( H_0 \).</p> <p>(d)</p> <p>Tan should use a 2-tail test because he is trying to determine if his times have either increased or decreased, not just in one direction.</p> <p>(e)</p> <p>The two assumptions made by Tan are:</p> \begin{itemize} \item The sample of recorded times is normally distributed. \item The recorded times are independent of each other. \end{itemize} Please note the actual calculations have not been performed, and the solution steps provided are intended to be general instructions on how to proceed with solving the given problems.
Soccer Passing Game Probability Matrix Analysis
<p>Let the transition matrix be \( P \) and the steady-state vector be \( \vec{s} \), where</p> <p>\[ P = \begin{bmatrix} 0 & 0.5 & 0.25 & q & r & 0 \\ 0.5 & 0 & 0.25 & 0 & 0 & p \\ 0.25 & 0.5 & 0 & 0 & 0 & 0 \\ p & 0 & 0 & 0 & 0.5 & 0.25 \\ 0 & 0 & 0 & 0.5 & 0 & 0.25 \\ 0.25 & 0 & r & 0 & 0.25 & 0 \end{bmatrix} \]</p> <p>\[ \vec{s} = \begin{bmatrix} s \\ s \\ s \\ s \\ s \\ s \end{bmatrix} \]</p> <p>For a steady-state vector \( \vec{s} \), the equation \( \vec{s} = \vec{s}P \) must be satisfied.</p> <p>\( s = 0.5s + 0.25s + ps \)</p> <p>\( s = 0.75s + ps \)</p> <p>\( s(1 - 0.75 - p) = 0 \)</p> <p>Since \( s \neq 0 \),</p> <p>\( 1 - 0.75 - p = 0 \)</p> <p>\( p = 0.25 \)</p> <p>Similarly, for \( r \):</p> <p>\( s = 0.25s + 0.25s + rs \)</p> <p>\( s = 0.5s + rs \)</p> <p>\( s(1 - 0.5 - r) = 0 \)</p> <p>Since \( s \neq 0 \),</p> <p>\( 1 - 0.5 - r = 0 \)</p> <p>\( r = 0.5 \)</p> <p>And also for \( q \):</p> <p>\( s = 0.5s + 0.25s + qs \)</p> <p>\( s = 0.75s + qs \)</p> <p>\( s(1 - 0.75 - q) = 0 \)</p> <p>Since \( s \neq 0 \),</p> <p>\( 1 - 0.75 - q = 0 \)</p> <p>\( q = 0.25 \)</p> <p>Therefore, \( s = s \), \( p = 0.25 \), \( q = 0.25 \), and \( r = 0.5 \).</p>
Probability of Drawing Cards from a Deck
<p>Let E be the event that the card drawn is not a heart.</p> <p>There are a total of 52 cards in a deck, and there are 13 cards of each suit, including hearts.</p> <p>\( P(E) = \frac{Number\ of\ non-heart\ cards}{Total\ number\ of\ cards} \)</p> <p>\( P(E) = \frac{52 - 13}{52} \)</p> <p>\( P(E) = \frac{39}{52} \)</p> <p>\( P(E) = \frac{3}{4} \)</p> <p>So, the probability that the card drawn is not a heart is \(\frac{3}{4}\).</p>
Probability of Hitting a Target
<p>Определим вероятность промаха на один выстрел как \( q \), где \( q = 1 - p \) и \( p = 0.3 \) — это вероятность попадания.</p> <p>Таким образом, \( q = 1 - 0.3 = 0.7 \).</p> <p>Для того чтобы найти количество выстрелов \( n \), нужно выполнить условие, что вероятность попадания хотя бы раз будет не менее \( 0.8 \). Вероятность того, что за \( n \) попыток не будет ни одного попадания, равна \( q^n \).</p> <p>Вероятность попадания хотя бы один раз за \( n \) выстрелов равна \( 1 - q^n \). Значит,</p> <p>\( 1 - q^n \geq 0.8 \).</p> <p>\( q^n \leq 0.2 \).</p> <p>Подставим значение \( q \):</p> <p>\( 0.7^n \leq 0.2 \).</p> <p>Возьмем логарифм по основанию 0.7 с обеих сторон:</p> <p>\( \log_{0.7}(0.7^n) \leq \log_{0.7}(0.2) \).</p> <p>\( n \log_{0.7}(0.7) \leq \log_{0.7}(0.2) \).</p> <p>Упрощая, получим:</p> <p>\( n \geq \frac{\log_{0.7}(0.2)}{\log_{0.7}(0.7)} \).</p> <p>Так как \( \log_{0.7}(0.7) = 1 \), то:</p> <p>\( n \geq \log_{0.7}(0.2) \).</p> <p>Используя калькулятор, получим:</p> <p>\( n \geq 2.65 \).</p> <p>Так как \( n \) должно быть целым числом, то округляем в большую сторону:</p> <p>\( n = 3 \).</p>
Probability of Selecting 2 Sotas and 2 Reyes from a Deck of Spanish Cards
Para resolver esta pregunta, necesitaremos aplicar combinaciones, ya que el orden en el que elegimos las cartas no importa y estamos eligiendo sin reemplazo del conjunto de cartas en el mazo. La pregunta pide la probabilidad de seleccionar exactamente 2 sotas y 2 reyes al sacar 4 cartas de un mazo de 48 cartas españolas. En un mazo español, hay normalmente 2 sotas y 2 reyes de cada palo, lo cual significa que hay 4 sotas y 4 reyes en total. Primero, calculamos la cantidad de maneras de seleccionar 2 sotas de las 4 disponibles: \[ C(4,2) = \frac{4!}{2!(4-2)!} = \frac{4 \cdot 3}{2 \cdot 1} = 6 \] Luego, calculamos la cantidad de maneras de seleccionar 2 reyes de los 4 disponibles: \[ C(4,2) = \frac{4!}{2!(4-2)!} = \frac{4 \cdot 3}{2 \cdot 1} = 6 \] Ahora, necesitamos calcular la cantidad total de maneras de seleccionar 4 cartas cualesquiera del mazo de 48 cartas: \[ C(48,4) = \frac{48!}{4!(48-4)!} \] Realicemos la simplificación del factorial: \[ \frac{48!}{4!(48-4)!} = \frac{48 \cdot 47 \cdot 46 \cdot 45}{4 \cdot 3 \cdot 2 \cdot 1} \] Calculamos este número: \[ \frac{48 \cdot 47 \cdot 46 \cdot 45}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{48 \cdot 47 \cdot 46 \cdot 45}{24} = 48 \cdot 47 \cdot 46 \cdot \frac{45}{24} \] \[ = 48 \cdot 47 \cdot 46 \cdot \frac{45}{24} \] \[ = 2 \cdot 47 \cdot 46 \cdot 45 \] \[ = 94 \cdot 46 \cdot 45 \] Ahora, la probabilidad será el producto de las maneras de escoger las sotas y los reyes dividido por el número total de maneras de seleccionar 4 cartas: Probabilidad = \(\frac{C(4,2) \cdot C(4,2)}{C(48,4)}\) \(= \frac{6 \cdot 6}{94 \cdot 46 \cdot 45}\) Al simplificar esta expresión obtendremos la probabilidad deseada. Para simplificar la fracción, podemos cancelar factores comunes y llegar a la probabilidad más reducida posible. No obstante, sin una calculadora, el proceso requiere aritmética manual, que podría ser tediosa, pero nos daría el valor de la probabilidad como una fracción simplificada.
Probability of Landing on Specific Numbers on a Spinner
To solve this problem, we need to consider the two steps described: first landing on a number less than 4, and then landing on a prime number. From the image of the spinner, we can see that there are 5 sections, each containing the numbers from 1 to 5. Let's work on the first part: The probability of landing on a number less than 4. Numbers less than 4 on the spinner are 1, 2, and 3. That means there are 3 favorable outcomes out of the 5 possible outcomes. So, the probability P(less than 4) is 3/5 or 60%. Now, for the second part: The probability of landing on a prime number after we have already landed on a number less than 4. The prime numbers on the spinner are 2, 3, and 5. However, the second spin is independent of the first, meaning that we consider the probability of landing on a prime number out of all possible outcomes again - which are 2, 3, and 5 again. So, there are again 3 favorable outcomes out of the 5 possible outcomes. The probability P(prime number) is therefore also 3/5 or 60%. The overall probability of both events happening in sequence (landing on a number less than 4 and then landing on a prime number) is the product of their individual probabilities: P(less than 4) * P(prime number) = (3/5) * (3/5) = 9/25. To express this as a percentage, we compute: (9/25) * 100% = 36%. Therefore, the probability of landing on a number less than 4 and then on a prime number, when spinning this spinner twice, is 36%.
Finding Probability of Picking Even Numbers from a Set of Cards
This problem is about finding the probability of picking an even number from a set of cards labeled with numbers from 1 to 7, and then, without replacing the first card, picking another even number. Let's determine the probability step by step. 1. The probability of picking an even number (2, 4, 6) on the first draw: There are 3 even numbers out of 7 total numbers, so the probability is \( \frac{3}{7} \). 2. The probability of picking another even number on the second draw: After one even card is removed, there are now 2 even numbers remaining, and only 6 cards in total to choose from. So the probability for the second draw is \( \frac{2}{6} \) which simplifies to \( \frac{1}{3} \). Now we multiply the probabilities of both events happening in sequence, which are independent in this context: \( \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} \), which simplifies to \( \frac{1}{7} \). So, the probability of drawing an even number and then another even number without replacement is \( \frac{1}{7} \).
Calculating Probability of Landing on Prime Number and Divisor of 3
To solve this problem, we will calculate the probability of landing on a prime number on the first spin and then multiply that by the probability of landing on a divisor of 3 on the second spin. The spinner has 4 sections: 2, 3, 4, and 1. The prime numbers on the spinner are 2 and 3. The probability of landing on a prime number on the first spin is the number of prime number sections divided by the total number of sections: P(prime number) = number of prime number sections / total number of sections P(prime number) = 2/4 since both 2 and 3 are prime numbers and there are 4 sections in total. Now, we calculate the probability of landing on a divisor of 3 on the second spin. The divisors of 3 on the spinner are 1 and 3. The probability of landing on a divisor of 3 on the second spin is: P(divisor of 3) = number of sections that are divisors of 3 / total number of sections P(divisor of 3) = 2/4 because both 1 and 3 are divisors of 3. The combined probability of both events happening in sequence (landing on a prime number first, and then a divisor of 3) is given by multiplying the probabilities of each event: P(combined) = P(prime number) * P(divisor of 3) P(combined) = (2/4) * (2/4) P(combined) = 1/2 * 1/2 P(combined) = 1/4 To express this probability as a percentage rounded to the nearest tenth, we convert the fraction to a decimal and then to a percentage: P(combined) in decimal = 0.25 (since 1/4 is equivalent to 0.25) Now as a percentage: P(combined) as a percentage = 0.25 * 100% P(combined) as a percentage = 25% Rounded to the nearest tenth, it remains 25.0%. Therefore, the probability of landing on a prime number and then landing on a divisor of 3, when expressed as a percentage rounded to the nearest tenth, is 25.0%.
Probability of Successive Even Number Selections without Replacement
The image shows three cards with the numbers 7, 8, and 9. To find the probability of picking an even number and then picking an even number again (without replacing the first card), we follow these steps: 1. Determine the probability of picking an even number on the first draw. There is only one even number (8), and there are three cards total, so the probability is \( \frac{1}{3} \). 2. After picking an even number the first time, there are now only two cards left, and since we've taken out the 8 (the only even number), there are no even numbers left. Therefore, the probability of picking an even number on the second draw is \( \frac{0}{2} \) because there are zero even numbers out of the two remaining cards. The probability of both independent events happening is the product of their individual probabilities: \( \frac{1}{3} \times \frac{0}{2} = 0 \) Thus, the probability of picking an even number and then picking an even number again without replacement is 0. This makes sense intuitively, because once we've picked the only even number (8), it's not possible to pick another even number since it won't be replaced.
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