Final answer:
To determine the probability of all four children receiving the same type of fruit, calculate the individual probabilities for all apples and all oranges and add them together. For the unfair coin, set up binomial probability equations for obtaining 2 and 3 heads in 7 tosses and equate them to solve for the probability of a single toss landing heads.
Step-by-step explanation:
Probability of All Children Receiving the Same Fruit
To find the probability that all four children receive the same fruit, we have two scenarios: all receive apples or all receive oranges. Initially, there are 5 apples and 6 oranges, totaling 11 fruits. If the first fruit is an apple, the chance the next three are also apples is (4/10) * (3/9) * (2/8), since there are fewer apples and fewer total fruits each time. The probability for all apples is then (5/11) * (4/10) * (3/9) * (2/8). Similarly, if the first fruit is an orange, the probability the next three are also oranges is (5/10) * (4/9) * (3/8). The overall probability is the sum of the probabilities for all apples and all oranges.
Unfair Coin Probability
Given that a certain unfair coin has the same probability for 2 heads out of 7 tosses as for 3 heads out of 7 tosses, we can set up a binomial distribution formula for each case and equate them. Let the probability of heads be p and tails be q, with p + q = 1. Using the combination formula for choosing k successes (heads) out of n trials (tosses), which is C(n, k) * p^k * q^(n-k), we get two equations. By equating these equations, we can solve for p. Among the options given, we need to find which probability would create equal outcomes for 2 heads and 3 heads in 7 tosses. This requires solving the binomial probability equations.
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