Question

In a college algebra class there are 18 freshman and 17 sophomores. Eight of freshman are females and 13 of sophomores are male. Student is selected at random, probability of selecting a female?

Three cards are drawn from a deck without replacement, what is the probability of selecting a king, a nine and a ten?

A light bulb is advertised as lasting a 100 hours with a standard deviation of 1500 hours. Find the probability of buying a light bulb and it lasting a) less than 1250 hours , b) between 901 and 1000 hours c) more than 901 hours

Show all work and formulas

Answer 1

1) There are 18 freshman and 17 sophomores. 8 of females are freshman out of 18.and 13 of sophomores are male, therefore, there are (17-13)= 4 female sophomores. there are total (8+4) = 12 females. we can choose a random person in 35C1 ways. we can select a random female in 12C1 ways.

Probability of selecting a female ,

  

= 0.3429

2) Three cards are drawn from a deck without replacement. There are 52 cards in a deck. we know, out of these 52 cards there are 4 kings,4 nine, 4 ten.

Probability of selecting a king from 52 cards = = 4/52

Probability of selecting a nine from (52-1) = 51 cards is = 4/51

Probability of selecting a ten from (51-1) = 50 cards is = 4/50

Now , Probability of selecting a king , a nine, a ten:

= 0.0005.

3) A light bulb is advertised as lasting a 100 hours with a standard deviation 1500 hours. Hence light bulb follows normal distribution with mean( ) 100 hours and standard deviation (s) 1500 hours

we need to find z-score here,

let,

now, Z ~ normal (0,1)

a) less than 1250 hours

P(X<z)

= P(X < 1250)

=

= P (Z < 23/30)

= P ( Z < 0.7667)

=

= 0.7793

b) between 901 and 1000 hours.

P (901 < X < 1000)

= P(Z < 0.60) - P(Z < 0.534)

= 0.7257469 - 0.7019440

= 0.0238

c) more than 901 hours

P(X > 901)

= 1 - 0.7019440

= 0.2980

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