A problem in mathematics is given to 4 students whose chances of solving individually are (1/2), (1/3), (1/4) and (1/5). probability that the problem will be solved at least by one student is

Question

A problem in mathematics is given to 4 students whose chances of solving individually are (1/2), (1/3), (1/4) and (1/5). probability that the problem will be solved at least by one student is (A) (2/3) (B) (3/5) (C) (4/5) (D) (3/4)

Tardigrade Admin · Accepted Answer

P(A ∪ B ∪ C ∪ D) = 1 -P(Ac ∩ Bc ∩ Cc ∩ Dc) = 1-P(Ac)⋅ P(Bc)⋅ P(Cc)⋅ P(Dc) =1-(1/2)×(2/3)×(3/4)×(4/5) =1-(1/5) = (4/5)

Q. A problem in mathematics is given to 4 students whose chances of solving individually are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5} .$ probability that the problem will be solved at least by one student is

$\frac{2}{3}$

$\frac{3}{5}$

$\frac{4}{5}$

$\frac{3}{4}$

$P (A \cup B \cup C \cup D) = 1 - P (A^{c} \cap B^{c} \cap C^{c} \cap D^{c})$
$= 1 - P (A^{c}) \cdot P (B^{c}) \cdot P (C^{c}) \cdot P (D^{c})$
$= 1 - \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}$
$= 1 - \frac{1}{5} = \frac{4}{5}$

Q. A problem in mathematics is given to 4 students whose chances of solving individually are 21​,31​,41​ and 51​. probability that the problem will be solved at least by one student is

32​

53​

54​

43​