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In successive rolls of a pair of fair dice, what is the probability of getting 2sevens before 6even numbers?

Short Answer

Expert verified

The probability of getting 2 sevens before 6 even numbers are55.5%

Step by step solution

01

Given Information

what is the probability of getting 2 sevens before even numbers

02

Step 2: Explanation

The probability that two sevens will be rolled before six-event numbers in a sequence of rolls of two dice

If any number is rolled that is not 7or even, it will be dismissed because it doesn't effect the probability in question

Name

C - 7or even number is rolled

S - a 7is rolled

E - an even number is rolled

03

Step 3: Explanation

P(C)

There are 36equally likely results of a roll of two dice: Twelve of which do not sum up to 7or even number-

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

The remaining 24form events c

P(C)=2436=23

P(S)

Six of those 36equally likely events are that the sum is 7

P(S)=636

P(E)

Since C=E∪S

P(E)=1836

Now the definition of conditional independence yields.

PC(S)=P(S∣C)=P(SC)P(C)=S⊆CP(S)P(C)=624=14

PC(E)=P(E∣C)=P(EC)P(C)=E⊆CP(E)P(C)=1824=34

04

Explanation

After sevenCrolls, either6even numbers or2sevens have been rolled. And precisely one of those happened.

The probability that two sevens were rolled before six even numbers is the probability that they have been rolled in the first 7important rolls.

Compute first the probability of the complement.

PC("six or seven even numbers are rolled" )=PC("6even numbers are rolled" )+PC("7even numbers are rolled" )

=7[PC(E)]6PC(S)+[PC(E)]7

=(34)6[7â‹…14+34]

=36â‹…1047

PC("at least two sevens are rolled")=1−PC("six or seven even numbers are rolled")

=1−36⋅1047

=55.5%

05

Final Answer

The probability of getting 2sevens before 6even numbers are=55.5%

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Most popular questions from this chapter

A and B flip coins. A starts and continues flipping

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