Question

if tossing two dice, whats the chance you do not get a sum of 7 or a 3 on either die?
○ 9/36
○ 20/36
○ 21/36
○ 16/36

Explanation:

Step1: Find total outcomes

When tossing two dice, each die has 6 faces. So total outcomes = \(6\times6 = 36\).

Step2: Find outcomes with sum 7

The pairs that sum to 7 are \((1,6)\), \((2,5)\), \((3,4)\), \((4,3)\), \((5,2)\), \((6,1)\). So there are 6 outcomes.

Step3: Find outcomes with a 3 on either die

For the first die being 3: \((3,1)\), \((3,2)\), \((3,3)\), \((3,4)\), \((3,5)\), \((3,6)\) (6 outcomes). For the second die being 3 (excluding those already counted where first die is 3): \((1,3)\), \((2,3)\), \((4,3)\), \((5,3)\), \((6,3)\) (5 outcomes). Total for a 3 on either die: \(6 + 5 = 11\). But we have overlapping outcomes (where sum is 7 and has a 3: \((3,4)\), \((4,3)\)) which are counted twice. So using inclusion - exclusion, outcomes with sum 7 or a 3 on either die: \(6+11 - 2=15\).

Step4: Find favorable outcomes (not sum 7 or 3 on either die)

Favorable outcomes = total outcomes - outcomes with sum 7 or 3 on either die = \(36-15 = 21\).

Step5: Find the probability

Probability = \(\frac{\text{favorable outcomes}}{\text{total outcomes}}=\frac{21}{36}\).

Answer:

21/36 (the option with 21/36)