If-the-probability-of-A-solving-a-question-is-1-2-and-the-probability-of-B-solving-the-question-is-2-3-then-the-probability-of-the-question-being-solved-is-

Question Number 210832 by BaliramKumar last updated on 19/Aug/24

$$$$If the probability of A solving a question is 1/2 and the probability of B solving the question is 2/3 then the probability of the question being solved is

Commented by mr W last updated on 20/Aug/24

$$bothAandB{don}^{\prime }tsolve:$$$$p=(1-\frac{1}{2})\times (1-\frac{2}{3})=\frac{1}{6}$$$$atleastoneofthemsolves:$$$$p=1-\frac{1}{6}=\frac{5}{6}✓$$

Answered by Ghisom last updated on 20/Aug/24

$$\mathrm{the}\mathrm{probability}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{solved}\mathrm{by}\mathrm{both}\mathrm{A}$$$$\mathrm{and}\mathrm{B}\mathrm{is}(1-\frac{1}{2})(1-\frac{2}{3})=\frac{1}{6}\Rightarrow \mathrm{the}$$$$\mathrm{probability}\mathrm{it}\mathrm{is}\mathrm{solved}\mathrm{is}1-\frac{1}{6}=\frac{5}{6}$$

Answered by BHOOPENDRA last updated on 20/Aug/24

$$$$The probability of the question being solved is the probability of at least one of them solving it. We can use the formula:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

where P(A ∪ B) is the probability of the question being solved, P(A) is the probability of A solving it, P(B) is the probability of B solving it, and P(A ∩ B) is the probability of both A and B solving it.

Assuming A and B are independent, we have:

P(A ∩ B) = P(A) ×P(B) = (1/2) × (2/3) = 1/3

Now, we can plug in the values:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= (1/2) + (2/3) – (1/3)
= 1/2 + 1/3
= 5/6

So, the probability of the question being solved is 5/6.
To find the probability of the question being unsolved, we can use the complement rule:

P(unsolved) = 1 – P(solved)

We already found the probability of the question being solved:

P(solved) = 5/6

So, the probability of the question being unsolved is:

P(unsolved) = 1 – 5/6
= 1/6

Therefore, the probability of the question being unsolved is 1/6.

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