Question

As an experiment two dies are rolled in two occasions. What is the probability of getting the numbers 8 or 7 in both occassions?
What is the probability of getting the numbers 9 and 10 in both occassions?

Answer 1

Correct Question-

As an experiment two dies are rolled in two occasions.

• What is the probability of getting the numbers sum as 8 or 7 in both occassions?
• What is the probability of getting the sum as 9 and 10 in both occassions?

Solution-

(i)

Total number of events when a dice is thrown twice is 36

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

Number of favorable events for getting sum as 7 is 6,P(A)=$\frac{6}{36}$

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

Number of favorable events for getting sum as 8 is 5,P(B)=$\frac{5}{36}$

»((6,2),(5,3),(4,4),(3,5),(2,6))

The probability of getting sum as either 7 or 8 becomes sum of individual probabilities of getting sum as 7 or 8

since there are no common favorable events their probability becomes 0

»P(AnB)=0

(P(AUB)=P(A)+P(B)-P(AnB))=(6/36)+(5/36)+0)=11/36

____________________________

(ii)

Number of favorable events for getting sum as 9 is 6,P(A)=$\frac{4}{36}$

»((5,4),(4,5),(3,6),(6,3))

»Number of favorable events for getting sum as 10 is 3,P(B)=$\frac{3}{36}$

»((5,5),(6,4),(4,6))

The probability of getting sum as either 7 or 8 becomes sum of individual probabilities of getting sum as 7 or 8

since there are no common favorable events their probability becomes 0

»P(AnB)=0

(P(AUB)=P(A)+P(B)-P(AnB))=(4/36)+(3/36)+0)=7/36

Answer 2

♧︎︎︎ Answer ♧︎︎︎

➪Total number of events when a dice is thrown twice is 36

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

• Number of favorable events for getting sum as 7 is 6,P(A)=\frac{6}{36}

• »((6,1),(5,2),(4,3),(2,5),(3,4),(1,6))
• Number of favorable events for getting sum as 8 is 5,P(B)=\frac{5}{36}

• »((6,2),(5,3),(4,4),(3,5),(2,6))
• The probability of getting sum as either 7 or 8 becomes sum of individual probabilities of getting sum as 7 or 8

• since there are no common favorable events their probability becomes 0

• »P(AnB)=0

• (P(AUB)=P(A)+P(B)-P(AnB))=(6/36)+(5/36)+0)=11/36

➪Number of favorable events for getting sum as 9 is 6,P(A)=\frac{4}{36}

• »((5,4),(4,5),(3,6),(6,3))

• »Number of favorable events for getting sum as 10 is 3,P(B)=\frac{3}{36}

• »((5,5),(6,4),(4,6))

The probability of getting sum as either 7 or 8 becomes sum of individual probabilities of getting sum as 7 or 8

• since there are no common favorable events their probability becomes 0

• »P(AnB)=0

• (P(AUB)=P(A)+P(B)-P(AnB))=(4/36)+(3/36)+0)=7/36

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