For numbers 1 to 4: Tay-Sachs (TS) disease is a rare genetic disorder passed from parents to children. Suppose the chance that a child will inherit the disease is 0.05 and a couple with such case has four children. Let X be the number of children with the disease. 3. TRUE or FALSE: The probability that none of the children will have the disease is 0.8145 O TRUE O FALSE
The correct answer and explanation is:
Answer: TRUE
Explanation:
In this case, we are dealing with a binomial probability scenario. The probability of a child inheriting Tay-Sachs disease is 0.05, and the chance that a child does not inherit the disease is 1 – 0.05 = 0.95.
The couple has four children, and we are asked to find the probability that none of the children will inherit the disease. This situation can be modeled using the binomial distribution, where:
- The number of trials (children) is n=4n = 4,
- The probability of success (inheriting the disease) is p=0.05p = 0.05,
- The probability of failure (not inheriting the disease) is q=1−p=0.95q = 1 – p = 0.95,
- We are looking for the probability of zero successes (children inheriting the disease), or X=0X = 0.
The formula for the binomial probability is: P(X=k)=(nk)pkqn−kP(X = k) = \binom{n}{k} p^k q^{n-k}
Where:
- P(X=k)P(X = k) is the probability of kk successes (children inheriting the disease),
- (nk)\binom{n}{k} is the binomial coefficient, representing the number of ways to choose kk successes from nn trials,
- pkp^k is the probability of kk successes, and
- qn−kq^{n-k} is the probability of n−kn-k failures.
For k=0k = 0 (none of the children inherit the disease), we substitute into the formula: P(X=0)=(40)(0.05)0(0.95)4P(X = 0) = \binom{4}{0} (0.05)^0 (0.95)^4 P(X=0)=1×1×(0.95)4P(X = 0) = 1 \times 1 \times (0.95)^4 P(X=0)=0.8145P(X = 0) = 0.8145
Thus, the probability that none of the four children will inherit the disease is 0.8145, which is the correct answer.
This calculation shows that there is a relatively high probability (81.45%) that none of the four children will inherit Tay-Sachs disease, given the low probability of 0.05 for any individual child.