Three dice are thrown randomly. The probability of coming 3 in at least one die is
91/216
Understanding the Probability of Coming 3 with Three Dice
The question asks for the probability of getting a 3 on at least one die when three dice are thrown randomly. This is a common type of probability problem that can be solved effectively by considering the complement event.
Total Possible Outcomes When Rolling Three Dice
When a single die is rolled, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). Since three dice are thrown, the total number of possible outcomes is the product of the outcomes for each die.
Total outcomes = Outcomes on die 1 $\times$ Outcomes on die 2 $\times$ Outcomes on die 3
Total outcomes = $6 \times 6 \times 6 = 216$
So, there are 216 equally likely outcomes when rolling three dice.
Finding the Probability of the Complement Event
The event "getting a 3 in at least one die" is the opposite or complement of the event "getting no 3s on any of the three dice". It's often easier to calculate the probability of the complement event and subtract it from 1.
Let's consider the complement event: getting no 3s on any of the three dice.
For a single die, the number of outcomes that are *not* a 3 is 5 (1, 2, 4, 5, 6).
If we want no 3s on any of the three dice, each die must show one of these 5 outcomes.
Number of outcomes with no 3s = Outcomes not 3 on die 1 $\times$ Outcomes not 3 on die 2 $\times$ Outcomes not 3 on die 3
Number of outcomes with no 3s = $5 \times 5 \times 5 = 125$
The probability of getting no 3s on any of the three dice is the number of outcomes with no 3s divided by the total number of outcomes.
Probability (no 3s) = $\frac{\text{Number of outcomes with no 3s}}{\text{Total number of outcomes}} = \frac{125}{216}$
Calculating the Probability of Coming 3 in at Least One Die
Now that we have the probability of the complement event (no 3s), we can find the probability of the original event (at least one 3) using the formula:
Probability (at least one 3) = 1 - Probability (no 3s)
Probability (at least one 3) = $1 - \frac{125}{216}$
To subtract, we find a common denominator:
Probability (at least one 3) = $\frac{216}{216} - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$
Thus, the probability of coming 3 in at least one die is $\frac{91}{216}$.
Summary of Calculation
| Event | Number of Outcomes | Probability |
|---|---|---|
| Total Outcomes (Three Dice) | $6^3 = 216$ | - |
| No 3s on any die | $5^3 = 125$ | $\frac{125}{216}$ |
| At least one 3 on a die | $216 - 125 = 91$ | $1 - \frac{125}{216} = \frac{91}{216}$ |
This calculation shows the steps to arrive at the probability of coming 3 on at least one die when three dice are thrown.
Important Questions from Probability
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