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A die is thrown three times. The probabi...

A die is thrown three times. The probability of obtaining at least one six is

A

`125/216`

B

`91/216`

C

`1/216`

D

`215/216`

Text Solution

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The correct Answer is:
To find the probability of obtaining at least one six when a die is thrown three times, we can follow these steps: ### Step 1: Understand the Problem We need to calculate the probability of getting at least one six in three throws of a die. **Hint:** Remember that "at least one" means we can have one, two, or three sixes. ### Step 2: Use the Complement Rule Instead of calculating the probability of getting one, two, or three sixes directly, we can use the complement rule. The complement of getting at least one six is getting no sixes at all. **Hint:** The complement rule states that P(A) = 1 - P(A'), where A' is the complement of event A. ### Step 3: Calculate the Probability of No Sixes The probability of not getting a six in a single throw of a die is \( \frac{5}{6} \). If the die is thrown three times, the probability of not getting a six in all three throws is: \[ P(\text{no sixes}) = \left(\frac{5}{6}\right)^3 \] **Hint:** When calculating the probability of independent events, multiply the probabilities of each event. ### Step 4: Calculate \( \left(\frac{5}{6}\right)^3 \) Now, we compute: \[ \left(\frac{5}{6}\right)^3 = \frac{5^3}{6^3} = \frac{125}{216} \] **Hint:** Make sure to calculate \( 5^3 \) and \( 6^3 \) correctly. ### Step 5: Use the Complement to Find the Desired Probability Now, we can find the probability of getting at least one six: \[ P(\text{at least one six}) = 1 - P(\text{no sixes}) = 1 - \frac{125}{216} \] **Hint:** Subtract the probability of the complement from 1. ### Step 6: Calculate the Final Probability Now, we perform the subtraction: \[ P(\text{at least one six}) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216} \] ### Step 7: Convert to Decimal (if needed) To express this probability as a decimal: \[ \frac{91}{216} \approx 0.4213 \] ### Final Answer Thus, the probability of obtaining at least one six when a die is thrown three times is approximately \( 0.421 \). ---
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Knowledge Check

  • An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

    A
    `(63)/(64)`
    B
    `(255)/(256)`
    C
    `(127)/(128)`
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    `(1)/(2)`
  • If a die is thrown 7 times, then the probability of obtaining 5 exactly 4 times is

    A
    `""^7C_(4)(1/6)^4(5/6)^3`
    B
    `""^7C_(4)(1/6)^3(5/6)^4`
    C
    `(1/6)^4(5/6)^3`
    D
    `(1/6)^3(5/6)^4`
  • Two dice thrown n times in succession The probability of obtaining a double six at least once is

    A
    `(1/36)^n`
    B
    `1-(35/36)^n`
    C
    `(1/12)^n`
    D
    None of these
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