Math  /  Data & Statistics

Question4. When rolling two dice, the probability of rolling doubles is 1/61 / 6. Suppose that a game player rolls the dice four times, hoping to roll doubles. Find the probability that the player gets doubles twice in 4 attempts.
1 2 3 4 11.6\% 1.6\% 98.4\% 88.4\% Angelina Piazza

Studdy Solution

STEP 1

What is this asking? If we try rolling two dice four times, what are the chances we get doubles exactly twice? Watch out! Don't forget that we can get doubles on *any* two of the four rolls, not just the first two!

STEP 2

1. Define the probability of success and failure
2. Set up the binomial probability formula
3. Calculate the binomial coefficient
4. Calculate the probability

STEP 3

Let pp be the probability of rolling doubles, which is p=16p = \frac{1}{6}.
This is our **probability of success**!

STEP 4

Let qq be the probability of *not* rolling doubles.
Since the total probability is 11, we have q=1p=116=56q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}.
This is our **probability of failure**!

STEP 5

We are given that the number of trials is n=4n = 4, and we want to find the probability of getting doubles exactly k=2k = 2 times.

STEP 6

The binomial probability formula gives the probability of exactly kk successes in nn trials: P(X=k)=(nk)pkqnkP(X=k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} Where (nk)\binom{n}{k} is the binomial coefficient, representing the number of ways to choose kk successes out of nn trials.

STEP 7

In our case, we have n=4n = 4, k=2k = 2, p=16p = \frac{1}{6}, and q=56q = \frac{5}{6}.
Let's plug these values into the formula: P(X=2)=(42)(16)2(56)42P(X=2) = \binom{4}{2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{4-2}

STEP 8

The binomial coefficient (42)\binom{4}{2} is calculated as: (42)=4!2!(42)!=4!2!2!=4321(21)(21)=244=6\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1)(2 \cdot 1)} = \frac{24}{4} = 6 This means there are **6 different ways** to get doubles exactly twice in four rolls!

STEP 9

Now, let's plug the binomial coefficient back into our probability formula: P(X=2)=6(16)2(56)2P(X=2) = 6 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^2

STEP 10

Calculate the powers: P(X=2)=61362536P(X=2) = 6 \cdot \frac{1}{36} \cdot \frac{25}{36}

STEP 11

Multiply the fractions: P(X=2)=6251296=1501296P(X=2) = 6 \cdot \frac{25}{1296} = \frac{150}{1296}

STEP 12

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 6: P(X=2)=1501296=25216P(X=2) = \frac{150}{1296} = \frac{25}{216}

STEP 13

Convert to a percentage: P(X=2)=252160.1157=11.57%11.6%P(X=2) = \frac{25}{216} \approx 0.1157 = 11.57\% \approx 11.6\%

STEP 14

The probability of getting doubles exactly twice in four attempts is approximately 11.6%11.6\%.

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