A die is rolled 4 times .What is the probability of getting only one 6?
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Probability of getting a six in one throw / roll : 1/6
We get one 6 only in four rolls or throws of the die. This means that the other three throws we get other than 6.
Probability of getting a non-six (from 1 to 5) in one throw or roll : 5/6
Throwing of a die first time is independent from other throws. So probabilities multiply.
You can get one 6 in the first throw, or 2nd throw, or 3rd throw or the fourth roll. So,
Probability = 1/6 * 5/6 * 5/6 * 5/6 + 5/6 * 1/6 * 5/6 * 5/6
+ 5/6 * 5/6 * 1/6 * 5/6 + 5/6 * 5/6 * 5/6 * 1/6
= 4 * 1* 5³ / 6⁴ = 0.3858
= 38.58 %
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For those who know binomial expansion :
(p + q)⁴ = gives probabilities for occurrences of 6 or other than 6 when a die is rolled 4 times. p = 1/6 q = 5/6. The second term in the expansion gives:
⁴C₁ p q³ = ⁴C₁ (1/6) (5/6)³ = 0.3858
We get one 6 only in four rolls or throws of the die. This means that the other three throws we get other than 6.
Probability of getting a non-six (from 1 to 5) in one throw or roll : 5/6
Throwing of a die first time is independent from other throws. So probabilities multiply.
You can get one 6 in the first throw, or 2nd throw, or 3rd throw or the fourth roll. So,
Probability = 1/6 * 5/6 * 5/6 * 5/6 + 5/6 * 1/6 * 5/6 * 5/6
+ 5/6 * 5/6 * 1/6 * 5/6 + 5/6 * 5/6 * 5/6 * 1/6
= 4 * 1* 5³ / 6⁴ = 0.3858
= 38.58 %
==============================================
For those who know binomial expansion :
(p + q)⁴ = gives probabilities for occurrences of 6 or other than 6 when a die is rolled 4 times. p = 1/6 q = 5/6. The second term in the expansion gives:
⁴C₁ p q³ = ⁴C₁ (1/6) (5/6)³ = 0.3858
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