A shipment of ten smartphones contains five with cracked screens. If sold in a

random order, what is the probability that at most three of the first six sold have

cracked screens?

73.81%

82.468%

41.259%

91.375%

Answer: 73.81%

Step-by-step explanation:

We know that 5 out of 10 have broken screens. So each smartphone has a probability of 50% of having a broken screen.

The probability that at most 3 of them have a broken screen is:

The probability that none has a broken screen p = 0

This is 0 because we are selling six of them and only 5 of them have the screen in good condition.

The probabilities are calculated as:

P = A*6! / (6 - n) !*n!

where n is the number of broken phones.

the probability is equal to

if we have only one broken screen:

for the broken phone, we have a 5/10 probability.

for the other 5 nice phones, the probabilities will be 5/9. 4/8. 3/7, 2/6 and 1/5 (the number changes because when we select one of the phones, the total number of phones decreases)

A = (5*5*4*3*2*1) / (10*9*8*7*6*5)

and is similar thinking for 2 and 3 broken phones.

Probability that one has a broken screen:

P = ((5*4*3*2*1*5) / (10*9*8*7*6*5)) * 6 = 0.0238

Probability for two broken screen:

P = (5*4*3*2*5*4) / (10*9*8*7*6*5)) * 6*5/2 = 0.2381

For 3 broken phones the probability is:

P = ((5*4*3*5*4*3) / (10*9*8*7*6*5)) * 6*5*4 / (2*3) = 0.4762

total probability = 0.0238 + 0.2381 + 0.4762 = 0.7381

or 73.81% in percentage form.