The probability of getting a total of $9$ exactly twice in $6$ tosses of a pair of dice is
- $_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9} \right) }^{ 2}+_{ 3 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 3 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 3 }$
- $_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
- $1-_{ 2 }^{ 6 }{ C }{ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
- ${ \left( \cfrac { 8 }{ 9 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 9 } \right) }^{ 2 }$
Reveal answer
Fill a bubble to check yourself
B
Correct answer
Explanation
Probability of getting 9 with two dice: (3,6), (4,5), (5,4), (6,3) = 4/36 = 1/9. Probability of not getting 9 = 8/9. Binomial distribution for 6 tosses, 2 successes: 6C2 * (1/9)^2 * (8/9)^4.
AI explanation
When two dice are tossed, there are 36 possible outcomes and 4 ways to get a sum of 9, so the probability of getting a 9 is 4/36 = 1/9 and the probability of failure is 8/9. Using the binomial distribution formula for exactly 2 successes in 6 trials, the probability is given by choosing 2 out of 6 trials, multiplied by the success probability squared and the failure probability to the fourth power. This yields 6C2 multiplied by (1/9)^2 and (8/9)^4.