Multiple choice

The chance that a person with two dices, the faces of each being numbered $1$ to $6$, will throw aces exactly $4$ times in $6$ trials is

  1. ${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }$
  2. ${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$
  3. $_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$
  4. $_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 2 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 4 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Getting aces with two dice means rolling two sixes, which has probability 1/36. Exactly four successes in six trials have probability C(6,4)(1/36)^4(35/36)^2, so option C is correct.

AI explanation

Using the binomial distribution formula, the probability of obtaining aces exactly 4 times in 6 trials requires the combination 6C4. The probability of success is throwing a double ace, which is 1/36, and the probability of failure is 35/36. Therefore, the final probability is 6C4 multiplied by (1/36)^4 and (35/36)^2.

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