The probability of selection $$= \dfrac{\text{No of favourable outcomes}}{\text{No of total possible outcomes}}$$
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The probability of selection $$= \dfrac{\text{No of favourable outcomes}}{\text{No of total possible outcomes}}$$
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If two dice are rolled, what is the probability that the sum of the faces is going to be even?
Sum of the faces would be even if both die are even or both are odd. Probability that both the faces are odd is $$\frac{1}{2} * \frac{1}{2}$$ Probability that both the faces are even is $$\frac{1}{2} * \frac{1}{2}$$ So P(E) = $$\frac{1}{2} * \frac{1}{2} +\frac{1}{2}*\frac{1}{2} = \frac{1}{2}$$.
If two dice are rolled, what is the probability that at least one of them is going to be even?
The probability that neither of the faces show an even number is 1/2 *1/2 =1/4. Hence the probability at least one is even is 1-1/4 = 0.75.
If a card is drawn randomly from a deck, what is the probability that the card would be neither a red card nor a face card?
Favourable outcome is if the card is a black card from 1 to 10. There are twenty such cards out of 52. Hence p = 20/52 = 5/13.
Mahesh wishes to bet on a horse race in such a way that he wins if either of his two picks comes in first. The odds in favour of his two picks are 3:17 and 1:9. What is the probability of Mahesh winning?
The probability of first horse winning = 3 / (3 +17) = 0.15. Probability of second horse is 1/(1+9)=0.1. As they are mutually exclusive, probability of Mahesh winning = 0.15+0.1= 0.25.
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