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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # One letter is selected at random from the 5 letters V, W, X, Y, and Z,

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Intern  B
Joined: 27 Aug 2018
Posts: 11
Location: India
GMAT 1: 670 Q51 V29 GPA: 4
WE: Analyst (Other)
One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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1
5 00:00

Difficulty:   75% (hard)

Question Stats: 49% (02:25) correct 51% (02:21) wrong based on 57 sessions

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One letter is selected at random from the 5 letters V, W, X, Y, and Z, and event A is the event that the letter V is selected. A fair six-sided die with sides numbered 1, 2, 3, 4, 5 and 6 is to be rolled, and event B is the event that a 5 or a 6 shows. A fair coin is to be tossed, and event C is the event that a head shows.

What is the probability that event A occurs and at least one of the events b and c occurs

A. 1/30
B. 1/15
C. 1/10
D. 2/15
E. 1/5
Math Expert V
Joined: 02 Aug 2009
Posts: 8758
One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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rakshitsood21 wrote:
One letter is selected at random from the 5 letters V, W, X, Y, and Z, and event A is the event that the letter V is selected. A fair six-sided die with sides numbered 1, 2, 3, 4, 5 and 6 is to be rolled, and event B is the event that a 5 or a 6 shows. A fair coin is to be tossed, and event C is the event that a head shows.

What is the probability that event A occurs and at least one of the events b and c occurs

A. 1/30
B. 1/15
C. 1/10
D. 2/15
E. 1/5

One letter is selected at random from the 5 letters V, W, X, Y, and Z, and event A is the event that the letter V is selected. -- $$\frac{1}{5}$$
A fair six-sided die with sides numbered 1, 2, 3, 4, 5 and 6 is to be rolled, and event B is the event that a 5 or a 6 shows. --$$\frac{2}{6}=\frac{1}{3}$$
A fair coin is to be tossed, and event C is the event that a head shows.--$$\frac{1}{2}$$

Now we are looking for 'the probability that event A occurs and at least one of the events b and c occurs'
Probability that both A and B occur - $$\frac{1}{5}*\frac{1}{3}$$
Probability that both A and B occur - $$\frac{1}{5}*\frac{1}{2}$$
Probability that all three A, B and C occur - $$\frac{1}{5}*\frac{1}{2}*\frac{1}{3}$$...Now this is part of both the above probabilities.

Answer = $$\frac{1}{5}(\frac{1}{2}+\frac{1}{3}-\frac{1}{2}*\frac{1}{3})=\frac{1}{5}*\frac{2}{3}=\frac{2}{15}$$

D
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Intern  B
Joined: 01 Sep 2019
Posts: 29
Concentration: Marketing, Strategy
Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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chetan2u wrote:
rakshitsood21 wrote:
One letter is selected at random from the 5 letters V, W, X, Y, and Z, and event A is the event that the letter V is selected. A fair six-sided die with sides numbered 1, 2, 3, 4, 5 and 6 is to be rolled, and event B is the event that a 5 or a 6 shows. A fair coin is to be tossed, and event C is the event that a head shows.

What is the probability that event A occurs and at least one of the events b and c occurs

A. 1/30
B. 1/15
C. 1/10
D. 2/15
E. 1/5

One letter is selected at random from the 5 letters V, W, X, Y, and Z, and event A is the event that the letter V is selected. -- $$\frac{1}{5}$$
A fair six-sided die with sides numbered 1, 2, 3, 4, 5 and 6 is to be rolled, and event B is the event that a 5 or a 6 shows. --$$\frac{2}{6}=\frac{1}{3}$$
A fair coin is to be tossed, and event C is the event that a head shows.--$$\frac{1}{2}$$

Now we are looking for 'the probability that event A occurs and at least one of the events b and c occurs'
Probability that both A and B occur - $$\frac{1}{5}*\frac{1}{3}$$
Probability that both A and B occur - $$\frac{1}{5}*\frac{1}{2}$$
Probability that all three A, B and C occur - $$\frac{1}{5}*\frac{1}{2}*\frac{1}{3}$$...Now this is part of both the above probabilities.

Answer = $$\frac{1}{5}(\frac{1}{2}+\frac{1}{3}-\frac{1}{2}*\frac{1}{3})=\frac{1}{5}*\frac{2}{3}=\frac{2}{15}$$

D

Hi chetan2u

Why did you subtract all three events together?
Intern  B
Joined: 09 May 2018
Posts: 8
Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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Hi chetan2u , did not understand the substraction part. Kindly help
Math Expert V
Joined: 02 Aug 2009
Posts: 8758
Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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Aman1204 and Kunni
When we are looking at A and B happening together, it includes when C happens with them.
Similarly when we are looking at A and C happening together, it includes When B happens with them.
So when you add both the above cases, all three happening together is added two times, so we have to subtract it once from the total.
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Manager  S
Joined: 05 Jan 2020
Posts: 129
Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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we have to find the probability of A and (B or C).

Now B or C means 1 - (!B and !C) = 1 - [(2/3)*(1/2)] = 2/3.

multiplying by probability of A we get 2/15.
Intern  S
Joined: 12 May 2020
Posts: 13
Location: India
GPA: 4
WE: Analyst (Consulting)
Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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Hi chetan2u,
There is a typo:
Probability that both A and B occur - 1/5∗1/2

Probability that both A and C occur - 1/5∗1/2
Senior Manager  P
Joined: 21 Jun 2017
Posts: 438
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,  [#permalink]

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P(required) = P(A)*(1- P(not B)*P(not C)) = 1/5 * (1 - [4/6 * 1/2]) = 2/15 ---> D Re: One letter is selected at random from the 5 letters V, W, X, Y, and Z,   [#permalink] 26 May 2020, 07:02

# One letter is selected at random from the 5 letters V, W, X, Y, and Z,   