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When throwing 2 dice together, all possible outcomes for a total of 4 are (1,3), (2,2), (3,1).
E = {♣J,àJ,J,♠J} There are four successful outcomes out of the possible 52. Therefore
$p(E)=\frac{4}{52}=\frac{1}{13}$
E = {♣J,àJ,J,♠J} There are four successful outcomes out of the possible 52. Therefore $p(E)=\frac{4}{52}=\frac{1}{13}$
One defective microprocessor and three good microprocessors are required to choose. We have 10 defective microprocessors and 90 (=100 - 10) good microprocessors, so the successful outcomes. Since any:
Since any four microprocessors can be chosen in C(100,4) ways, so:
$p(E)=\frac{C(10,1)\times C(90,3)}{C(100,4)}\simeq 0.2996$