Since we are replacing the cards, the trials are independent. So this is a binom
ID: 2930172 • Letter: S
Question
Since we are replacing the cards, the trials are independent. So this is a binomial distribution Since r A were added to the pack, there are now (52 +x) cards in pack out of which (1+x) are the AO. So, in 10 attempts, the expected value is: 1+1=032 * (52 +z) r*(1-0.32) = 52 *032-1 0.682 = 15.64 r = 23 So, there are 23 extra Acards in the pack Variance for n independent trials for binomial distribution is: = n * P * (1-P) = 10 * 0.32 * (1-032) = 10 * 032 * 0.68 = 2.176 (b) Since we are replacing the card, this is a binomial distribution. We are getting less than 2 AC. The sample space will contain 5110 cases of O A as we can choose from 51 cards on each of the 10 trials. Also, the sample space will contain 10 24*51 cases of A as we can choose from 51 cards for each of the 9 trials and 24 cards for the remaining trial. There are 10 such combinations for the 10 trials. 10 5124 240 80 240 +10,51 +24 = 51 +240 = 291 = 97 =waa (c) This is a geometric distribution. For geometric distribution: E0323.125 (d) This is a geometie distribution. P(Not getting A on fir#t three trials) = (1-P = (1-032) = 0.68, 0.3144 (e) This is a binomial distribution. P(Getting 2nd card on 5th attempt) =P(Getting 18t card in 1st 4 att )-P(Getting 2nd card on 5th attempt) = (4 * 0.32+ 0.683) * 0.32 = 4 * 0.32, 0.68, 0.1288Explanation / Answer
Answer to the question is as follows:
a. Correct. We have applied the fact that in an binomial distribution Expected value =3.2 = np
b. Correct.
c. Correct
d. Correct. We have used the complementry method in probability to solve this problem
e. Correct. The calculation is 4C1*.32^1(.68^3) <-- 1st term. & 2nd term is .32 ( simple probability of getting the card)