A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.

2 answers:

8 0

Answer:

1/32

15/32

Step-by-step explanation:

For a fair sided coin,

Probability of heads, P(H) = 1/2

Probability of tails P(T) = 1/2

For a coin tossed 5 times,

P( All heads)

= P(HHHHH),

= P (H) x P(H) x P(H) x P(H) x P(H)

= (1/2) x (1/2) x (1/2) x (1/2) x (1/2)

= 1/32 (Ans)

For part B, it is easier to just list the possible outcomes for

"at most 2 heads" aka "could be 1 head" or "could be 2 heads"

"One Head" Outcomes:

P(HTTTT), P(THTTT) P(TTHTT), P(TTTHT), P(TTTTH)

"2 Heads" Outcomes:

P(HHTTT), P(HTHTT), P(HTTHT), P(HTTTH), P(THHTT), P(THTHT), P(THTTH), P(TTHHT), P(TTHTH), P(TTTHH)

If we count all the possible outcomes, we get 15 possible outcomes representing "at most 2 heads)

we know that each outcome has a probability of 1/32

hence 15 outcomes for "at most 2 heads" have a probability of

(1/32) x 15 = 15/32

6 0

$|\Omega|=2^5=32$

$|A|=1\\P(A)=\dfrac{1}{32}=3.125\%$

$|A|=1+5+10=16\\P(A)=\dfrac{16}{32}=\dfrac{1}{2}=50\%$

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3 years ago

DENIUS [597]

Answer:

Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

a) $T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0)$ . These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).

b) $T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5)$ . Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).