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erica [24]
3 years ago
14

A person has a 35 percent chance of winning on a scratch off lottery ticket. What is the probability she first wins of he sixth

ticket?
A.(0.65)^3(0.35)
B.(6/3)(0.65)^3(0.35)
C.(0.35)^3(0.65)
D, (6/1)(0.35)^3(0.65)
E. 0.35
Mathematics
1 answer:
Rudiy273 years ago
3 0

None of the options are correct

Step-by-step explanation:

Given :

a person has 35% chance of winning on a scratch- off lottery ticket.

To find the probability that she wins  the sixth ticket is,

P( she first wins the sixth ticket )

= _ _ _ _ _ _

the first 5 dashes represent does not win values ( prob = 0.65)

the last dash represents the win ( prob = 0.35)

= (0.65) ^5 * 0.35                          {assuming independence

none of the option is correct.

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(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

Step-by-step explanation:

For this case when a pair of dice is rolled we have the following sample pace for the outcomes:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

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For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

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(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

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