A die used in a certain board game has eight faces, of which 3 are red, 3 are yellow, and 2 are blue. Each face is equally likely to land faceup when the die is tossed. In the game, a player tosses the die until blue lands faceup, and the number of tosses before blue lands faceup is counted. For example, a player who tosses the sequence shown in the following table has tossed the die 3 times before blue lands faceup. What is the probability that a player will toss the die at least 2 times before blue lands faceup? A 0.1406 B 0.4219 C 0.4375 D 0.5625 E 0.5781

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

60.9774

step-by-step explanation:

because when you add 47.6 plus 7.32 equals 54.92

and when you add .0574 + 6= 6.0574

and when you add 54.92 and 6.0574 you get the answer of 60.9774.

plzz mark brainlest last one got deleted. cause i didn't explain it but i did but i understand what the mod meant

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof

answer: 12/52

step-by-step explanation: my reasoning is that there are 12 face cards in a deck and a total of 52 cards in all.