Question

48. Find the probability that a randomly selected student is male, given that the student's preference was Harry Potter.: * Oa. 0.3600 Ob. 0.2941 Oc. 0.6400 Od. 0.8170 Questions 49-50: If we pick a year at random and count how many Friday 13th's are in the year, it turns out there will be 1, 2, or 3 such days. (In other words, every year has to have at least one Friday 13th but no more than three. It can be shown that the probability a randomly chosen year has one, two, or three Friday 13th's is as follows: # Friday 13th's P(x) XP(x) 1 0.4275 2 0.425 3 0.1475 49. Find the probability that a year selected at random has at least two Friday 13th's.:* O a. 0.1475 Ob. 0.425 Oc.0.5725 Od 0.575 50. In a given year, how many Friday 13th's do we expect to see?: * O a. 1.51 Ob. 1.72 Oc. 1.77 Od. 2.01

Answer 1

Solution :

49) Probability of an event E is given as follows :

P(E) = Number of favourable outcomes/Total number of outcomes

We have to find P(at least 2 firday 13th's).

P(at least 2 Friday 13th's) = P(2 Friday 13th's) + P(3 Friday 13th's)

P(at least 2 Friday 13th's) = 0.425 + 0.1475

P(at least 2 Friday 13th's) = 0.5725

Hence, the required probability of 0.5725.

Option (c) is correct.

50) The expected value is given as follows :

$\large E(X) = \sum X.P(x)$

# Friday 13th's	P(x)	X.P(x)
1	0.4275	0.4275
2	0.425	0.850
3	0.1475	0.4425
	Total	1.72

From the above table we have,

$\large \sum X.P(x) = 1.72$

$\large \therefore E(X) = 1.72$

Option (b) is correct.

​​​​​Note : For question number 48, please provide the data. The given information is not sufficient to solve question 48.