A fair die is tossed and the number facing up is noted. If the probability of getting at least one ‘six’ is to exceed 0.9, how many times should the die be tossed.
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To determine how many times the die should be tossed in order for the probability of getting at least one 'six' to exceed 0.9, we can use the concept of the complement.
The complement in this case is the probability of not getting any 'six' in a single toss of the die. Since each face of the die has an equal probability of occurring (1/6), the probability of not getting a 'six' in a single toss is 5/6.
We can now calculate the probability of not getting any 'six' in n tosses of the die. Since each toss is independent, the probability for each toss is multiplied together. So, the probability of not getting any 'six' in n tosses is (5/6)^n.
Now, we want the probability of getting at least one 'six' to exceed 0.9. In other words, we want the probability of not getting any 'six' in n tosses to be less than or equal to 0.1 (1 - 0.9 = 0.1).
So, (5/6)^n ≤ 0.1
To find the value of n that satisfies this inequality, we can take the natural logarithm of both sides:
ln((5/6)^n) ≤ ln(0.1)
n * ln(5/6) ≤ ln(0.1)
Now, we can solve for n:
n ≥ ln(0.1) / ln(5/6)
Using a calculator, we can find that ln(0.1) / ln(5/6) ≈ 25.19.
Since we can't have a fraction of a toss, we need to round up to the nearest whole number. Therefore, the die should be tossed at least 26 times in order for the probability of getting at least one 'six' to exceed 0.9.