An experiment results in one of the sample points \( E_{1}, E_{2}, E_{3}, E_{4} \), or \( E_{5} \). Complete parts a through \( c \). a. Find \( P\left(E_{3}\right) \) if \( P\left(E_{1}\right)=0.2, P\left(E_{2}\right)=0.2, P\left(E_{4}\right)=0.2 \), and \( P\left(E_{5}\right)=0.2 \). \( P\left(E_{3}\right)=\square \) (Type an exact answer in simplified form.) b. Find \( P\left(E_{3}\right) \) if \( P\left(E_{1}\right)=P\left(E_{3}\right), P\left(E_{2}\right)=0.2, P\left(E_{4}\right)=0.2 \), and \( P\left(E_{5}\right)=0.1 \). \( P\left(E_{3}\right)=\square( \) Type an exact answer in simplified form.) c. Find \( P\left(E_{3}\right) \) if \( P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{4}\right)=P\left(E_{5}\right)=0.2 \).

Someone hands you a box of 13 chocolate-covered candies, telling you that 6 are vanilla creams and the rest are peanut butter. You pick candies at random and discover the first 3 you eat are all vanilla. Consider a result to be surprising if there is less than a \( 5 \% \) chance of observing it given your expectations. Complete parts a through c below. a) If there really were 6 vanilla and 7 peanut butter candies in the box, what is the probability that you would have picked 3 vanillas in a row? (Round to three decimal places as needed.)

Topic - Probability A restaurant offers a 3 course meal. The entrée may be garlic bread \( (G) \) or chicken soup (S), the a) Draw a tree diagram and list (sample space) all the possible combinations for a 3 course meal.

En una bolsa hay cuatro tarjetas, con los números \( 1,2,3 \) y 4 respectivamente. Si se sacan dos tarjetas de la bolsa, una tras otra y con reposición, ¿cuál es la probabilidad de que la suma de los números obtenidos sea 7 ? A. \( \frac{1}{4} \) B. \( \frac{7}{16} \) C. \( \frac{1}{16} \)

9.1 Given: \( P(A)=0,45 ; P(B)=y \) and \( P(A \) or \( B)=0,74 \). Determine the value(s) of \( y \) : 9.1.1 If \( A \) and \( B \) are mutually exclusive events. 9.1.2 If \( A \) and \( B \) are independent events.

The probability that it will be suminy tomorrow is \( \frac{2}{9} \), If it is sunny, the probability tha Siya plays rugby is \( \frac{8}{15} \). If it is not sumy, the probability that Stya plays rugby is \( \frac{4}{15} \) 9.2.1 Represent the above information on the tree diagram and indicate all the possible outcomes.