Anjali mam test 2

.docx

School

Algoma University *

*We aren’t endorsed by this school

Course

101

Subject

Mathematics

Date

Apr 3, 2024

Type

docx

Pages

5

Uploaded by SargentFog19998 on coursehero.com

NAME – NIKHIL STUDENT ID – A00177149 TEACHER NAME – ANJALI VERMA TEST 2 1.A single 6-sided die is rolled. What is the probability of rolling a 1 or a prime number? Answer :- The prime numbers from 1 to 6 = 2, 3, and 5. favorable outcomes: favorable outcome of rolling 1 = 1 favorable outcome of rolling prime no. = 3 favorable outcomes together: 1 + 3 = 4. Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes) Total no. of possible outcome when roll a six side dice = 6 P = NFO/NPO PROBABILITY = 4/6 =2/3 2. A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a king or a club? Answer :- the probability of choosing a king : - 4/52
The probability of choosing a club :- 13/52 =1/4 the probability of choosing both a king and a club is 1/52. Probability of (King or Club) = (Probability of King) + (Probability of Club) - (Probability of both King and Club) = (4/52) + (13/52) - (1/52) = (4 + 13 - 1) / 52 = 16/52 =4/13 So, the probability of choosing a king or a club from a standard deck of 52 playing cards is 4/13. 3. Toss four coins. Let X be the number of heads showing. Find the mean and variance.( 2 points) Answers :- Mean (Expected Value) of X :- First we have to calculate the expected value of x There are two possible outcome when we tossing a coin , head or tail Probability of getting head = 1/2 Probability of getting tail =1/2 Now, for four coin tosses: • The probability of getting 0 heads (TTTT) = (1/2)^4 = 1/16 • The probability of getting 1 head (HTTT, THTT, TTHT, TTTH) = 4 * (1/2)^4 = 4/16 = 1/4 • The probability of getting 2 heads (HHTT, HTHH, THHT, TTHH, HTTH, THTH) = 6 * (1/2)^4 = 6/16 = 3/8
• The probability of getting 3 heads (HHHT, HTHH, THHH, HHTH) = 4 * (1/2)^4 = 4/16 = 1/4 • The probability of getting 4 heads (HHHH) = (1/2)^4 = 1/16 Mean (expected value): E(X) = (0 * 1/16) + (1 * 1/4) + (2 * 3/8) + (3 * 1/4) + (4 * 1/16) E(X) = 0 + 1/4 + 3/4 + 3/4 + 0 E(X) = 7/4 So, the mean (expected value) of X is 7/4 let's find the variance of X: Variance of X: To find the variance, we should calculate E(X^2) - [E(X)]^2. First, calculate E(X^2) using the probabilities and values of X: E(X^2) = (0^2 * 1/16) + (1^2 * 1/4) + (2^2 * 3/8) + (3^2 * 1/4) + (4^2 * 1/16) E(X^2) = (0 * 1/16) + (1/4) + (3/2) + (3/4) + 1 E(X^2) = 7/4 Now, calculate the variance: Var(X) = E(X^2) - [E(X)]^2 Var(X) = (7/4) - (7/4)^2 Var(X) = (7/4) - (49/16) Var(X) = (28/16) - (49/16) Var(X) = -21/16 So, the variance of X is -21/16. 4. Let X = number of prior convictions for prisoners at a state prison at which there are 500 prisoners. (x=0,1,2,3,4,5)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help