![]() Thus, the total probability of all seven events is 7/8. ![]() The probability of each of these seven ways is equal to 1/8. So, there are seven possible ways that Michael can toss at least one head. ![]() Because each coin toss is independent, we can multiply the probabilities together.įor example, the probability of the combination HTT is (1/2)(1/2)(1/2) = 1/8 The probability of rolling a head is ½ and the probability of rolling a tail is ½. We must find the probability of each of these ways and then add them together. Thus, there are seven ways that Michael can toss at least one head. If Michael tosses three heads, then there is only one possible combination: If Michael tosses two heads, then there are three possible combinations: We could model it like this, where H represents heads and T represents tails. If Michael tosses one head, then it could be on either the first, second, or third toss. Michael can toss either one head, two heads, or three heads. G - (13 - g) > 3 (Distribute negative sign in parentheses)Ģg - 13 > 3 (Add 13 to both sides of the inequality)Ģg > 16 (Divide both sides of the inequality by 2) ![]() You have b + g = 13 and g - b > 3, where b and g are positive integers.ī + g = 13 (Subtract g on both sides of the equation) You take the difference in marbles, which is 3, which means you need the difference in green and blue marbles to be greater than 3, or at least 4. Therefore, we need to take at least 9 green marbles, which would mean 4 or less of the marbles would be blue (8 green and 5 blue would leave us with equal green and equal blue marbles, so it would have to be more than 8 green marbles, which gives us 9 green marbles). If there are going to be more blue than green marbles after John's 13 marbles, he has to take at least 4 more green marbles than blue marbles, because right now there are 3 less blue marbles. John then takes 3 green and 6 blue from the bowl. This gives us 27 green and 27 blue marbles: The bowl has 54 marbles, half green and half blue. ![]()
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