Answer to Question #236785 in Statistics and Probability for Lee3415

Question

6. How many 4-digit odd numbers >4000 can be formed using the numbers 0,2,3,4,5,8 with 2 used exactly once without repetition of digits?

7. A bag consists of 4 red, 3 green, 2 white and 1 purple marbles. 5 are randomly selected. Find the probability of selecting 2 reds and 1 of each of the other colours.

8. A fair die is rolled 10 times.
Determine the probability that a 6 is displayed thrice.

9. 5 cards are dealt, without replacement, from 52-pack set of cards.
Find the probability that a distinct 5-card poker
hand will display at most 2 DIAMONDS.

10. A MCQ consists of 10 questions each having 4 options. Find probability that the student answers at least 3 questions correctly

Answer 1

6. How many 4-digit odd numbers >4000 can be formed using the numbers 0,2,3,4,5,8 with 2 used exactly once without repetition of digits?

We use product rule:

1-place may be 4,5,8-3 possibilities

In each case position for 2 may be of 2,3,4 without constraints or 3 possibilities

For other two        places we have 4 digits and therefore 4∙3=12 opportunities without repetition

Finally according  to product rule there are N=3∙3∙12=108 cases.

7. A bag consists of 4 red, 3 green, 2 white and 1 purple marbles. 5 are randomly selected. Find the probability of selecting 2 reds and 1 of each of the other colours.

For all combiтations there are

$C_{10}^5=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{1\cdot 2\cdot 3\cdot 4\cdot 5}=2\cdot 3\cdot 2\cdot 7\cdot 6=504$

For desired combination we hаve

$С_4^2\cdot C_3^1\cdot C_2^1\cdot C_1^1=\\ \frac{4\cdot 3}{1\cdot 2}\cdot 3\cdot 2\cdot 1=36$ cases

Thus desired probabality equals to $\frac{36}{504}=\frac{2}{27}$

8. A fair die is rolled 10 times.
determine the probability that a 6 is displayed thrice

At whole there are N=$6^{10}$ possibilities

Cases with two 6 there are $C_{10}^2=\frac{10\cdot 9}{1\cdot 2}\cdot 5^8=45\cdot 5^8$

Thus probability of the situation is $\frac{45\cdot 5^8}{6^{10}}=0.75\cdot (\frac{5}{6})^8$

9. 5 cards are dealt, without replacement, from 52-pack set of cards.

Find the probability that a distinct 5-card poker

hand will display at most 2 DIAMONDS.

Solution:

N₁=$C_{39}^5=\frac{39!}{5!34!}-$ cases with 0 diamonds

$N_2=13\cdot C_{39}^4=\frac{13\cdot 39!}{4!\cdot 35!}$ cases with exactly 1 diamond

$N_3=C_{13}^{2}\cdot C_{39}^{3}=\frac{108\cdot 39!}{3!36!}$ cases with two diamonds

$N=C_{52}^5=\frac{52!}{5!47!}$ - number of all possible cases

$p=\frac{N_1+N_2+N_3}{N}=\frac{\frac{39!}{5!34!}+\frac{13\cdot 39!}{4!\cdot 35!}+\frac{108\cdot 39!}{3!36!}}{\frac{52!}{5!47!}}$ =

=(35∙…∙47)/(40∙…52)+(13∙5∙36∙…47)/(40∙…52)+(108∙20∙37∙..47) /(40∙…52)=(35∙36+65∙36+108∙20)∙(37/42)∙(38/41)∙…∙(47/52)/(40∙41)

10. A MCQ consists of 10 questions each having 4 options. Find probability that the student answers at least 3 questions correctly

Solution:

We use the fornula $P(n,k)=C_n^k\cdot p^k\cdot (1-p)^{n-k}$ for having exactly k right answers with probability p for each. Thus

$P=1-P(10,0)-P(10,1)-P(10,2)=1-((\frac{3}{4})^{10}+C_{10}^1\cdot \frac{1}{4}\cdot (\frac{3}{4})^9+ C_{10}^2\cdot (\frac{1}{4})^2\cdot(\frac{3}{4})^8)=\\C_{10}^3\cdot (\frac{1}{4})^3\cdot (\frac{3}{4})^7=\\ =1-(\frac{3}{4})^8\cdot ((\frac{3}{4})^2+10\cdot \frac{1}{4}\cdot (\frac{3}{4})+ 45\cdot (\frac{1}{4})^2)$