Probability Multiplication Rule in Aptitude:

Overview:

Multiplication Rule of Probability:

If the probability of winning \(A\) is \(x\) and the probability of winning \(B\) is \(y\), then the probability of winning \(A\) AND \(B\) both.$$ \left[ P \ (A) \times P \ (B) \right] $$ where as,

\(P \ (A)\) = Probability of \(A\) to win.

\(P \ (B)\) = Probability of \(B\) to win.

Multiplication Rule of Probability Examples:

Example (1): If the probability of \(p\), winning a game is \(\frac{5}{4}\) and the probability of \(q\), winning is \(\frac{4}{3}\), then the probability of winning \(p\) AND \(q\) both-

Solution: Given values, \(P \ (p) = \frac{5}{4}\) and \(P \ (q) = \frac{4}{3}\) then, $$ \left[ P \ (p) \times P \ (q) \right] $$ $$ = \left[ \frac{5}{4} \times \frac{4}{3} \right] = \frac{5}{3}$$

Example (2): If the probability of passing in an exam is \(\frac{2}{3}\) for \(x\) and the probability of passing the same exam is \(\frac{4}{3}\) for \(y\), then the probability of passing \(x\) AND \(y\) both will be-

Solution: Given values, \(P \ (x) = \frac{2}{3}\) and \(P \ (y) = \frac{4}{3}\) then, $$ \left[ P \ (x) \times P \ (y) \right] $$ $$ = \left[ \frac{2}{3} \times \frac{4}{3} \right] = \frac{8}{9}$$

Example (3): If the balls are not being replaced, then find the chance of drawing \(2\) Red balls in succession from a bag containing \(7\) Blue and \(3\) Red balls?

Solution: Probability of getting Red balls in first draw = \(\frac{3}{10}\)

Probability of getting Red balls in second draw = \(\frac{2}{9}\) then, $$ P = \left[ \frac{3}{10} \times \frac{2}{9} \right] = \frac{1}{15}$$