A random sample of 75 high schoolers was drawn. Of these 50 do both snow-boardin

Question

A random sample of 75 high schoolers was drawn. Of these 50 do both snow-boarding and down hill skiing, 40 do both down hill skiing and cross country skiing, 15 do only cross country skiing, and 30 do all three. Suppose further that those who snow-board either also ski downhill or do all three and those who ski down hill either do at least two of the activities or all three.

-i- Find the probability that a randomly selected high schooler is either a snowboarder or a down hill skier.

-ii- Find the probability that a randomly selected high schooler is either a cross country or a down hill skier.

-iii- Find the probability that a randomly selected high schooler is a either a cross country or down hill skier, but not a snow boarder.

Explanation / Answer

Here we are given that total number of people taken are 75. Therefore we have here:

P(S) = 75,

n( SB and DHS) + n(all 3 ) = 50 where SB is snow boarding and DHS is Down Hill skiing.

Note that we added all 3 here because if we are given that: 50 do both snow-boarding and down hill skiing, these would also contain people who do all 3 of the sports.

n( DHS and CCS) + n(all 3 ) = 40 where CCS is cross country skiing

n( CCS Only ) = 15

n( All 3 ) = 30

Now using n( All 3 ) = 30 , put this in n( SB and DHS) + n(all 3 ) = 50 to get:

n( SB and DHS) = 50 - 30 = 20

Similarly ,n( DHS and CCS) + n(all 3 ) = 40

Therefore, n( DHS and CCS) = 40 - 30 = 10

Next we are given that , those who snow-board either also ski downhill or do all three. This means that:

n ( SB and CCS ) = 0 and n( SB only ) = 0

Also we are given that : those who ski down hill either do at least two of the activities or all three.

which means that n ( DHS only ) = 0

Now the total sample size is 75. This is computed as:

n ( DHS only ) + n( SB only ) + n(CCS only ) + n ( SB and CCS ) + n( DHS and CCS) + n( SB and DHS) + n(all 3 )

Putting all the values we get:

0 + 0 + 15 + 0 + 10 + 20 + 30 = 75

Hence verified that we are correct about our numbers.

(i) Now computing the probability that a randomly selected high schooler is either a snowboarder or a down hill skier, is computed as:

= [ 75 - n(CCS only) ] / 75

= [ 75 - 15 ] / 75

= 60/75

= 0.8

Therefore 0.8 is the required probability here.

(ii) Now computing the probability that a randomly selected high schooler is either a cross country or a down hill skier.

= [ 75 - n(SB only) ] / 75

= [ 75 - 0 ] / 75

= 1

Therefore 1 is the required probability here.

(iii) Now the probability that a randomly selected high schooler is a either a cross country or down hill skier, but not a snow boarder is computed as:

= [ n(CCS only ) + n(DHS only ) + n ( CCS and DHS ) ] / 75

= [ 15 + 0 + 10 ] / 75

= 1 / 3

Therefore 0.3333 is the required probability here.

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