Probability

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New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

17. The sample space of the experiment is

S = {1, 2, 3, 4, 5, 6} and E = {4}

F = {2, 4, 6}

So, E∩ F = {4} ∩ {2, 4, 6} = {4} ≠

Therefore E and F are not mutually exclusive.

New answer posted

6 months ago

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alok kumar singh

Contributor-Level 10

16. When a die is drawn we can have 1, 2, 3, 4, 5 and 6. So, sample space of throwing dice until a six comes up is

S = { (6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (1, 6)……., (1, 5, 6), (1, 6)……… (1, 5, 6)………}

Hence, the sample space is indefinite.

New answer posted

6 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

15. The possible outcome when a coin is tossed is a head or a tail. When a dice is thrown we can have the possible outcomes 1, 2, 3, 4, 5 and 6. Let R1, R2 be the two red balls and B1, B2, B3 be the two black balls. So, the desired sample space is

S = { (T, R1), (T, R2), (T, B1), (T, B2), (T, B3), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

New answer posted

6 months ago

0 Follower 1 View

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alok kumar singh

Contributor-Level 10

14. The possible outcome when a coin is tossed is a head or a tail. When a die is thrown we can have the possible outcome 1, 2, 3, 4, 5 and 6. So, the desired sample space is

S = { (2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, H, H), (1, H, T), (1, T, H), (1, T, T), (3, H, H), (3, H, T), (3, T, H), (3, T, T), (5, H, H), (5, H, T), (5, T, H), (5, T, T)}

New answer posted

6 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

13. The possible numbers to be chosen are 1, 2, 3 and 4. The sample space of drawing two slips one after another without replacement is

S = { (1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

New answer posted

6 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

12. When a coin is thrown we have the possible outcome of a head or a tail. And when a dice is thrown we have the possible outcomes 1, 2, 3, 4, 5 and 6. So, the desired sample space is

S = {T, (H, 1), (H, 2, 1), (H, 2, 2), (H, 2, 3), (H, 2, 4), (H, 2, 5), (H, 2, 6), (H, 3), (H, 4, 1), (H, 4, 2), (H, 4, 3), (H, 4, 4), (H, 4, 5), (H, 4, 6), (H, 5), (H, 6, 1), (H, 6, 2), (H, 6, 3), (H, 6, 4), (H, 6, 5), (H, 6, 6)}

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

11. Let us denote the non-defective and defective bulbs by 'N' and 'D'. So, that the sample space of selecting 3 bulbs from a lot is.

S = {NNN, NND, NDN, DNN, DDN, NDD, DND, DDD}

New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

11. The possible outcomes when a coin is tossed is a Head or a Tail. And the possible outcomes when a die is rolled is 1, 2, 3, 4, 5 or 6.

By condition, when a head occurs or first toss the coin is tossed again and if a tail occurs on first toss a die is rolled. So, the required sample space is

S = { (H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5) and (T, 6)}

New answer posted

6 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

9. Let us denote the red and white balls by R and W. We are to draw one ball and then another ball without replacement. So, the desired sample space is

S = {RW, WR, WW}

New answer posted

6 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

8. Let us denote boys and girls by B and G.

(i) The desired sample space whether a boy or girl are the children according to their birth.

S = {BB, BG, GB, GG}

(ii) A family of two children can be two girls, one girl or no girl. So, the desired sample space is S = {0, 1, 2}

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