A recent survey of Canadian university students indicated that 1 in 4, or 25%, a

Question

A recent survey of Canadian university students indicated that 1 in 4, or 25%, are holders of at least two credit cards1. Suppose you are to randomly pick two Canadian university students. What is the probability that both will be holders of at least two credit cards? neither will be holders of at least two credit cards? at least one of the two will be holders of at least two credit cards? Suppose you are to randomly pick n-Canadian university students in such a way that the probability that at least one of them will hold two or more credit cards will be 0.95. What is the minimum number of Canadian university students you should pick? In other words, find the value of n.

Explanation / Answer

a)

P(both) = 0.25*0.25 = 0.0625

b)

P(neither) = (1-0.25)*(1-0.25) = 0.5625

c)

P(at least one) = 1 - P(neither) = 0.4375

d)

In general,

P(no one has at least two CC) = (1-0.25)^n = 0.75^n

Thus,

P(at least one) = 1 - 0.75^n = 0.95

Solving for n,

1 - 0.75^n = 0.95

0.75^n = 0.05

n = ln(0.05)/ln(0.75)

n = 10.41334362

Rounding up so that we are more sure that it is at least 0.95,

n = 11 [ANSWER]

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