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Bunuel
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We can solve this question easily using venn diagram
160 has exactly 1, therefore none = 40
Every SMS has Email alert, therefore only(SMS)=0
150 do not have SMS, therefore 50 should have SMS, which should be common in both SMS & Email
Remaining part will be only(Email)=110

P(SMS|Email)=50/160=5/16
P(neither)=40/200=1/5

Bunuel
A mobile app lets users enable Email alerts and SMS alerts. Any user who has SMS alerts also has Email alerts. In a user base of 200 accounts, exactly 160 accounts have at least one of these alerts enabled, and exactly 150 accounts do not have SMS alerts.

One account is selected at random. Select for P(SMS|Email) the probability that the account has SMS alerts given that it has Email alerts, and select for P(neither) the probability that the account has neither alert. Make only two selections, one in each column.
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Bunuel
A mobile app lets users enable Email alerts and SMS alerts. Any user who has SMS alerts also has Email alerts. In a user base of 200 accounts, exactly 160 accounts have at least one of these alerts enabled, and exactly 150 accounts do not have SMS alerts.

One account is selected at random. Select for P(SMS|Email) the probability that the account has SMS alerts given that it has Email alerts, and select for P(neither) the probability that the account has neither alert. Make only two selections, one in each column.

Total Accounts = 200

From the question stem, 4 combinations of alerts will be there
Any user who has SMS alerts also has Email alerts ---> N(SMS & No Email) = 0
[color=#0f0f0f]exactly 150 accounts do not have SMS alerts. -->N(No SMS) = 150 --->N(SMS) =50


Till now, diagram will be like
Total 200
├── SMS (50) │
├── Email (50) │[/color]
└── No Email (0)
└── No SMS (150)
├── Email
└── No Email

[color=#0f0f0f]exactly 160 accounts have at least one of these alerts enabled
N[
Atleast One Alert] =160
--> N[No SMS &no Email]=40
-->N[No SMS& Email] =N[No SMS] - N[No SMS &no Email] =150 -40 =110

Now Our diagram is complete
Total 200
├── SMS (50) │
├── Email (50) │
└── No Email (0)
└── No SMS (150)
├── Email (110)
└── No Email (40)



P(SMS|Email) the probability that the account has SMS alerts given that it has Email alerts -> Out of Email alerts ,how many are having SMS alerts
= N[SMS & Email] /Total Email
=50/110+50 = 50/160= 5/16

P(neither) the probability that the account has neither alert. -> out of total users, how many dont have both
= N[no SMS & no Email ] / Total Users
= 40/200 = 1/5


[/color]

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Bunuel
A mobile app lets users enable Email alerts and SMS alerts. Any user who has SMS alerts also has Email alerts. In a user base of 200 accounts, exactly 160 accounts have at least one of these alerts enabled, and exactly 150 accounts do not have SMS alerts.

One account is selected at random. Select for P(SMS|Email) the probability that the account has SMS alerts given that it has Email alerts, and select for P(neither) the probability that the account has neither alert. Make only two selections, one in each column.
Let’s consider a 4*4 matrix

1) Any user who has SMS alerts also has Email alerts.

If a user has SMS alerts also has email alerts. Then having sms and no email , becomes 0.

Thus c = 0.

2) User base of 200 accounts, alteast one of the alerts enabled = 160 accounts.

ATLEAST ONE = SMS only + Email only + both sms and email = 160

therefore NO SMS and NO Email becomes = (200 - 160) = 40

d = 40

3) Exactly 150 don’t have SMS, thus, b+ d = 150

with d = 40, then b = 150 - 40 = 110.

if b+d = 150, then (a+c) = 200 - 150 = 50

a+ c = 50.

we know that c = 0, then a = 50


SMS NO SMS TOTAL
EMAIL a = 50 b = 110 160
NO EMAIL c = 0 d = 40 40
TOTAL 50 150 200


P ( SMS/ Email) = 50/160 = 5/16

P ( Neither ) = 40/200 = 1/5
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